m着色问题|回溯5

给定一个无向图和一个数字m，请确定该图是否最多可以用m种颜色着色，以使该图的两个相邻顶点都不用同一颜色着色。在此，图形着色是指将颜色分配给所有顶点。

输入输出格式：

输入：

1. 二维数组graph [V] [V]，其中V是图形中的顶点数，而graph [V] [V]是图形的邻接矩阵表示。如果存在从i到j的直接边，则值graph [i] [j]为1，否则graph [i] [j]为0。
2. 整数m是可以使用的最大颜色数。

输出：
数组color [V]的数字应为1到m。 color [i]应该表示分配给第i个顶点的颜色。如果图表无法使用m种颜色上色，则代码也应返回false。

例子：

Input:  
graph = {0, 1, 1, 1},
        {1, 0, 1, 0},
        {1, 1, 0, 1},
        {1, 0, 1, 0}
Output: 
Solution Exists: 
Following are the assigned colors
 1  2  3  2
Explanation: By coloring the vertices 
with following colors, adjacent 
vertices does not have same colors

Input: 
graph = {1, 1, 1, 1},
        {1, 1, 1, 1},
        {1, 1, 1, 1},
        {1, 1, 1, 1}
Output: Solution does not exist.
Explanation: No solution exits.

以下是可以用3种不同颜色进行着色的图形示例。

我们强烈建议您单击此处并进行实践，然后再继续解决方案。

方法1：天真。

天真的方法：生成所有可能的颜色配置。由于每个节点都可以使用m种可用颜色中的任何一种进行着色，因此可能的颜色配置总数为m ^ V。
生成颜色配置后，检查相邻顶点是否具有相同的颜色。如果满足条件，请打印组合并中断循环。

算法：

1. 创建一个使用当前索引，顶点数和输出颜色数组的递归函数。
2. 如果当前索引等于顶点数。检查输出颜色配置是否安全，即检查相邻顶点是否没有相同颜色。如果满足条件，请打印配置并中断。
3. 为顶点分配颜色(1到m)。
4. 对于每一个分配的颜色递归调用与下一个索引和顶点的号码的函数
5. 如果任何递归函数返回true，则中断循环并返回true。

以下是上述想法的实现：

C++

#include
using namespace std;
 
// Number of vertices in the graph
#define V 4
 
void printSolution(int color[]);
 
// check if the colored
// graph is safe or not
bool isSafe(bool graph[V][V], int color[])
{
    // check for every edge
    for (int i = 0; i < V; i++)
        for (int j = i + 1; j < V; j++)
            if (graph[i][j] && color[j] == color[i])
                return false;
    return true;
}
 
/* This function solves the m Coloring
problem using recursion. It returns
false if the m colours cannot be assigned,
otherwise, return true and prints
assignments of colours to all vertices.
Please note that there may be more than
one solutions, this function prints one
of the feasible solutions.*/
bool graphColoring(bool graph[V][V], int m, int i,
                int color[V])
{
    // if current index reached end
    if (i == V) {
       
        // if coloring is safe
        if (isSafe(graph, color)) {
           
            // Print the solution
            printSolution(color);
            return true;
        }
        return false;
    }
 
    // Assign each color from 1 to m
    for (int j = 1; j <= m; j++) {
        color[i] = j;
 
        // Recur of the rest vertices
        if (graphColoring(graph, m, i + 1, color))
            return true;
 
        color[i] = 0;
    }
 
    return false;
}
 
/* A utility function to print solution */
void printSolution(int color[])
{
    cout << "Solution Exists:" " Following are the assigned colors \n";
    for (int i = 0; i < V; i++)
        cout << "  " << color[i];
    cout << "\n";
}
 
// Driver code
int main()
{
    /* Create following graph and
    test whether it is 3 colorable
    (3)---(2)
    | / |
    | / |
    | / |
    (0)---(1)
    */
    bool graph[V][V] = {
        { 0, 1, 1, 1 },
        { 1, 0, 1, 0 },
        { 1, 1, 0, 1 },
        { 1, 0, 1, 0 },
    };
    int m = 3; // Number of colors
 
    // Initialize all color values as 0.
    // This initialization is needed
    // correct functioning of isSafe()
    int color[V];
    for (int i = 0; i < V; i++)
        color[i] = 0;
 
    if (!graphColoring(graph, m, 0, color))
        cout << "Solution does not exist";
 
    return 0;
}
 
// This code is contributed by shivanisinghss2110

C

#include 
#include 
 
// Number of vertices in the graph
#define V 4
 
void printSolution(int color[]);
 
// check if the colored
// graph is safe or not
bool isSafe(bool graph[V][V], int color[])
{
    // check for every edge
    for (int i = 0; i < V; i++)
        for (int j = i + 1; j < V; j++)
            if (graph[i][j] && color[j] == color[i])
                return false;
    return true;
}
 
/* This function solves the m Coloring
   problem using recursion. It returns
  false if the m colours cannot be assigned,
  otherwise, return true and prints
  assignments of colours to all vertices.
  Please note that there may be more than
  one solutions, this function prints one
  of the feasible solutions.*/
bool graphColoring(bool graph[V][V], int m, int i,
                   int color[V])
{
    // if current index reached end
    if (i == V) {
        // if coloring is safe
        if (isSafe(graph, color)) {
            // Print the solution
            printSolution(color);
            return true;
        }
        return false;
    }
 
    // Assign each color from 1 to m
    for (int j = 1; j <= m; j++) {
        color[i] = j;
 
        // Recur of the rest vertices
        if (graphColoring(graph, m, i + 1, color))
            return true;
 
        color[i] = 0;
    }
 
    return false;
}
 
/* A utility function to print solution */
void printSolution(int color[])
{
    printf("Solution Exists:"
           " Following are the assigned colors \n");
    for (int i = 0; i < V; i++)
        printf(" %d ", color[i]);
    printf("\n");
}
 
// Driver program to test above function
int main()
{
    /* Create following graph and
       test whether it is 3 colorable
      (3)---(2)
       |   / |
       |  /  |
       | /   |
      (0)---(1)
    */
    bool graph[V][V] = {
        { 0, 1, 1, 1 },
        { 1, 0, 1, 0 },
        { 1, 1, 0, 1 },
        { 1, 0, 1, 0 },
    };
    int m = 3; // Number of colors
 
    // Initialize all color values as 0.
    // This initialization is needed
    // correct functioning of isSafe()
    int color[V];
    for (int i = 0; i < V; i++)
        color[i] = 0;
 
    if (!graphColoring(graph, m, 0, color))
        printf("Solution does not exist");
 
    return 0;
}

Java

public class GFG
{
 
  // Number of vertices in the graph
  static int V = 4;
 
  /* A utility function to print solution */
  static void printSolution(int[] color)
  {
    System.out.println("Solution Exists:"  +
                       " Following are the assigned colors ");
    for (int i = 0; i < V; i++)
      System.out.print("  " + color[i]);
    System.out.println();
  }
 
  // check if the colored
  // graph is safe or not
  static boolean isSafe(boolean[][] graph, int[] color)
  {
    // check for every edge
    for (int i = 0; i < V; i++)
      for (int j = i + 1; j < V; j++)
        if (graph[i][j] && color[j] == color[i])
          return false;
    return true;
  }
 
  /* This function solves the m Coloring
    problem using recursion. It returns
    false if the m colours cannot be assigned,
    otherwise, return true and prints
    assignments of colours to all vertices.
    Please note that there may be more than
    one solutions, this function prints one
    of the feasible solutions.*/
  static boolean graphColoring(boolean[][] graph, int m,
                               int i, int[] color)
  {
    // if current index reached end
    if (i == V) {
 
      // if coloring is safe
      if (isSafe(graph, color))
      {
 
        // Print the solution
        printSolution(color);
        return true;
      }
      return false;
    }
 
    // Assign each color from 1 to m
    for (int j = 1; j <= m; j++)
    {
      color[i] = j;
 
      // Recur of the rest vertices
      if (graphColoring(graph, m, i + 1, color))
        return true;
      color[i] = 0;
    }
    return false;
  }
 
  // Driver code
  public static void main(String[] args)
  {
     
    /* Create following graph and
        test whether it is 3 colorable
        (3)---(2)
        | / |
        | / |
        | / |
        (0)---(1)
        */
    boolean[][] graph = {
      { false, true, true, true },
      { true, false, true, false },
      { true, true, false, true },
      { true, false, true, false },
    };
    int m = 3; // Number of colors
 
    // Initialize all color values as 0.
    // This initialization is needed
    // correct functioning of isSafe()
    int[] color = new int[V];
    for (int i = 0; i < V; i++)
      color[i] = 0;
    if (!graphColoring(graph, m, 0, color))
      System.out.println("Solution does not exist");
  }
}
 
// This code is contributed by divyeh072019.

Python3

# Number of vertices in the graph
# define 4 4
 
# check if the colored
# graph is safe or not
def isSafe(graph, color):
   
    # check for every edge
    for i in range(4):
        for j in range(i + 1, 4):
            if (graph[i][j] and color[j] == color[i]):
                return False
    return True
 
# /* This function solves the m Coloring
# problem using recursion. It returns
# false if the m colours cannot be assigned,
# otherwise, return true and prints
# assignments of colours to all vertices.
# Please note that there may be more than
# one solutions, this function prints one
# of the feasible solutions.*/
def graphColoring(graph, m, i, color):
   
    # if current index reached end
    if (i == 4):
 
        # if coloring is safe
        if (isSafe(graph, color)):
 
            # Prthe solution
            printSolution(color)
            return True
        return False
 
    # Assign each color from 1 to m
    for j in range(1, m + 1):
        color[i] = j
 
        # Recur of the rest vertices
        if (graphColoring(graph, m, i + 1, color)):
            return True
        color[i] = 0
    return False
 
# /* A utility function to prsolution */
def printSolution(color):
    print("Solution Exists:" " Following are the assigned colors ")
    for i in range(4):
        print(color[i],end=" ")
 
# Driver code
if __name__ == '__main__':
   
    # /* Create following graph and
    # test whether it is 3 colorable
    # (3)---(2)
    # | / |
    # | / |
    # | / |
    # (0)---(1)
    # */
    graph = [
        [ 0, 1, 1, 1 ],
        [ 1, 0, 1, 0 ],
        [ 1, 1, 0, 1 ],
        [ 1, 0, 1, 0 ],
    ]
    m = 3 # Number of colors
 
    # Initialize all color values as 0.
    # This initialization is needed
    # correct functioning of isSafe()
    color = [0 for i in range(4)]
 
    if (not graphColoring(graph, m, 0, color)):
        print ("Solution does not exist")
 
# This code is contributed by mohit kumar 29

C#

using System;
class GFG {
     
    // Number of vertices in the graph
    static int V = 4;
     
    /* A utility function to print solution */
    static void printSolution(int[] color)
    {
        Console.WriteLine("Solution Exists:"  +
                          " Following are the assigned colors ");
        for (int i = 0; i < V; i++)
            Console.Write("  " + color[i]);
        Console.WriteLine();
    }
 
    // check if the colored
    // graph is safe or not
    static bool isSafe(bool[,] graph, int[] color)
    {
        // check for every edge
        for (int i = 0; i < V; i++)
            for (int j = i + 1; j < V; j++)
                if (graph[i, j] && color[j] == color[i])
                    return false;
        return true;
    }
     
    /* This function solves the m Coloring
    problem using recursion. It returns
    false if the m colours cannot be assigned,
    otherwise, return true and prints
    assignments of colours to all vertices.
    Please note that there may be more than
    one solutions, this function prints one
    of the feasible solutions.*/
    static bool graphColoring(bool[,] graph, int m,
                              int i, int[] color)
    {
        // if current index reached end
        if (i == V) {
            
            // if coloring is safe
            if (isSafe(graph, color)) {
                
                // Print the solution
                printSolution(color);
                return true;
            }
            return false;
        }
      
        // Assign each color from 1 to m
        for (int j = 1; j <= m; j++) {
            color[i] = j;
      
            // Recur of the rest vertices
            if (graphColoring(graph, m, i + 1, color))
                return true;
      
            color[i] = 0;
        }
      
        return false;
    }
 
  // Driver code 
  static void Main() {
    /* Create following graph and
    test whether it is 3 colorable
    (3)---(2)
    | / |
    | / |
    | / |
    (0)---(1)
    */
    bool[,] graph = {
        { false, true, true, true },
        { true, false, true, false },
        { true, true, false, true },
        { true, false, true, false },
    };
    int m = 3; // Number of colors
  
    // Initialize all color values as 0.
    // This initialization is needed
    // correct functioning of isSafe()
    int[] color = new int[V];
    for (int i = 0; i < V; i++)
        color[i] = 0;
  
    if (!graphColoring(graph, m, 0, color))
        Console.WriteLine("Solution does not exist");
  }
}
 
// this code is contributed by divyeshrabadiya07.

Javascript

     
    // Number of vertices in the graph
    let V = 4;
     
    /* A utility function to print solution */
    function printSolution(color)
    {
        document.write("Solution Exists:"  +
       " Following are the assigned colors <br>");
        for (let i = 0; i < V; i++)
              document.write("   " + color[i]);
        document.write(" ");
    }
     
    // check if the colored
  // graph is safe or not
    function isSafe(graph,color)
    {
        // check for every edge
    for (let i = 0; i < V; i++)
      for (let j = i + 1; j < V; j++)
        if (graph[i][j] && color[j] == color[i])
          return false;
    return true;
    }
     
     
  /* This function solves the m Coloring
    problem using recursion. It returns
    false if the m colours cannot be assigned,
    otherwise, return true and prints
    assignments of colours to all vertices.
    Please note that there may be more than
    one solutions, this function prints one
    of the feasible solutions.*/
    function graphColoring(graph,m,i,color)
    {
        // if current index reached end
    if (i == V) {
  
      // if coloring is safe
      if (isSafe(graph, color))
      {
  
        // Print the solution
        printSolution(color);
        return true;
      }
      return false;
      }
  
    // Assign each color from 1 to m
    for (let j = 1; j <= m; j++)
    {
      color[i] = j;
  
      // Recur of the rest vertices
      if (graphColoring(graph, m, i + 1, color))
        return true;
      color[i] = 0;
    }
    return false;
    }
     
    // Driver code
     
    /* Create following graph and
        test whether it is 3 colorable
        (3)---(2)
        | / |
        | / |
        | / |
        (0)---(1)
        */
    let graph=[[ false, true, true, true],
               [ true, false, true, false ],
               [ true, true, false, true ],
               [true, false, true, false]];
     
    let m = 3; // Number of colors
     
    // Initialize all color values as 0.
    // This initialization is needed
    // correct functioning of isSafe()
    let color = new Array(V);
    for (let i = 0; i < V; i++)
      color[i] = 0;
       
    if (!graphColoring(graph, m, 0, color))
      document.write("Solution does not exist");
     
    
    // This code is contributed by unknown2108
     

---

输出

Solution Exists: Following are the assigned colors 
  1  2  3  2

复杂度分析：

• 时间复杂度： O(m ^ V)。
  总共有O(m ^ V)种颜色组合。因此，时间复杂度为O(m ^ V)。
• 空间复杂度： O(V)。
  graphColoring(…)函数的递归堆栈将需要O(V)空间。

方法2 ：回溯。

方法：想法是从顶点0开始将颜色逐一分配给不同的顶点。在分配颜色之前，请考虑已为相邻顶点分配的颜色来检查安全性，即检查相邻顶点是否具有相同的颜色。 。如果有任何不违反条件的颜色分配，请将颜色分配标记为解决方案的一部分。如果无法进行颜色分配，则回溯并返回false。

算法：

1. 创建一个使用图，当前索引，顶点数和输出颜色数组的递归函数。
2. 如果当前索引等于顶点数。在输出数组中打印颜色配置。
3. 为顶点分配颜色(1到m)。
4. 对于每种分配的颜色，请检查配置是否安全(即检查相邻顶点是否具有相同的颜色)以下一个索引和顶点数递归调用该函数
5. 如果任何递归函数返回true，则中断循环并返回true。
6. 如果没有递归函数返回true，则返回false。

以下是上述想法的实现：

C++

#include 
#include 
 
// Number of vertices in the graph
#define V 4
 
void printSolution(int color[]);
 
/* A utility function to check if
   the current color assignment
   is safe for vertex v i.e. checks
   whether the edge exists or not
   (i.e, graph[v][i]==1). If exist
   then checks whether the color to
   be filled in the new vertex(c is
   sent in the parameter) is already
   used by its adjacent
   vertices(i-->adj vertices) or
   not (i.e, color[i]==c) */
bool isSafe(
    int v, bool graph[V][V],
    int color[], int c)
{
    for (int i = 0; i < V; i++)
        if (
            graph[v][i] && c == color[i])
            return false;
    return true;
}
 
/* A recursive utility function
to solve m coloring problem */
bool graphColoringUtil(
    bool graph[V][V], int m,
    int color[], int v)
{
    /* base case: If all vertices are
       assigned a color then return true */
    if (v == V)
        return true;
 
    /* Consider this vertex v and
       try different colors */
    for (int c = 1; c <= m; c++) {
        /* Check if assignment of color
           c to v is fine*/
        if (isSafe(
                v, graph, color, c)) {
            color[v] = c;
 
            /* recur to assign colors to
               rest of the vertices */
            if (
                graphColoringUtil(
                    graph, m, color, v + 1)
                == true)
                return true;
 
            /* If assigning color c doesn't
               lead to a solution then remove it */
            color[v] = 0;
        }
    }
 
    /* If no color can be assigned to
       this vertex then return false */
    return false;
}
 
/* This function solves the m Coloring
   problem using Backtracking. It mainly
   uses graphColoringUtil() to solve the
   problem. It returns false if the m
   colors cannot be assigned, otherwise
   return true and prints assignments of
   colors to all vertices. Please note
   that there may be more than one solutions,
   this function prints one of the
   feasible solutions.*/
bool graphColoring(
    bool graph[V][V], int m)
{
    // Initialize all color values as 0.
    // This initialization is needed
    // correct functioning of isSafe()
    int color[V];
    for (int i = 0; i < V; i++)
        color[i] = 0;
 
    // Call graphColoringUtil() for vertex 0
    if (
        graphColoringUtil(
            graph, m, color, 0)
        == false) {
        printf("Solution does not exist");
        return false;
    }
 
    // Print the solution
    printSolution(color);
    return true;
}
 
/* A utility function to print solution */
void printSolution(int color[])
{
    printf(
        "Solution Exists:"
        " Following are the assigned colors \n");
    for (int i = 0; i < V; i++)
        printf(" %d ", color[i]);
    printf("\n");
}
 
// driver program to test above function
int main()
{
    /* Create following graph and test
       whether it is 3 colorable
      (3)---(2)
       |   / |
       |  /  |
       | /   |
      (0)---(1)
    */
    bool graph[V][V] = {
        { 0, 1, 1, 1 },
        { 1, 0, 1, 0 },
        { 1, 1, 0, 1 },
        { 1, 0, 1, 0 },
    };
    int m = 3; // Number of colors
    graphColoring(graph, m);
    return 0;
}

Java

/* Java program for solution of
   M Coloring problem using backtracking */
public class mColoringProblem
{
    final int V = 4;
    int color[];
 
    /* A utility function to check
       if the current color assignment
       is safe for vertex v */
    boolean isSafe(
        int v, int graph[][], int color[],
        int c)
    {
        for (int i = 0; i < V; i++)
            if (
                graph[v][i] == 1 && c == color[i])
                return false;
        return true;
    }
 
    /* A recursive utility function
       to solve m coloring  problem */
    boolean graphColoringUtil(
        int graph[][], int m,
        int color[], int v)
    {
        /* base case: If all vertices are
           assigned a color then return true */
        if (v == V)
            return true;
 
        /* Consider this vertex v and try
           different colors */
        for (int c = 1; c <= m; c++)
        {
            /* Check if assignment of color c to v
               is fine*/
            if (isSafe(v, graph, color, c))
            {
                color[v] = c;
 
                /* recur to assign colors to rest
                   of the vertices */
                if (
                    graphColoringUtil(
                        graph, m,
                        color, v + 1))
                    return true;
 
                /* If assigning color c doesn't lead
                   to a solution then remove it */
                color[v] = 0;
            }
        }
 
        /* If no color can be assigned to
           this vertex then return false */
        return false;
    }
 
    /* This function solves the m Coloring problem using
       Backtracking. It mainly uses graphColoringUtil()
       to solve the problem. It returns false if the m
       colors cannot be assigned, otherwise return true
       and  prints assignments of colors to all vertices.
       Please note that there  may be more than one
       solutions, this function prints one of the
       feasible solutions.*/
    boolean graphColoring(int graph[][], int m)
    {
        // Initialize all color values as 0. This
        // initialization is needed correct
        // functioning of isSafe()
        color = new int[V];
        for (int i = 0; i < V; i++)
            color[i] = 0;
 
        // Call graphColoringUtil() for vertex 0
        if (
            !graphColoringUtil(
                graph, m, color, 0))
        {
            System.out.println(
                "Solution does not exist");
            return false;
        }
 
        // Print the solution
        printSolution(color);
        return true;
    }
 
    /* A utility function to print solution */
    void printSolution(int color[])
    {
        System.out.println(
            "Solution Exists: Following"
            + " are the assigned colors");
        for (int i = 0; i < V; i++)
            System.out.print(" " + color[i] + " ");
        System.out.println();
    }
 
    // driver program to test above function
    public static void main(String args[])
    {
        mColoringProblem Coloring
               = new mColoringProblem();
        /* Create following graph and
           test whether it is
           3 colorable
          (3)---(2)
           |   / |
           |  /  |
           | /   |
          (0)---(1)
        */
        int graph[][] = {
            { 0, 1, 1, 1 },
            { 1, 0, 1, 0 },
            { 1, 1, 0, 1 },
            { 1, 0, 1, 0 },
        };
        int m = 3; // Number of colors
        Coloring.graphColoring(graph, m);
    }
}
// This code is contributed by Abhishek Shankhadhar

Python

# Python program for solution of M Coloring
# problem using backtracking
 
class Graph():
 
    def __init__(self, vertices):
        self.V = vertices
        self.graph = [[0 for column in range(vertices)]\
                              for row in range(vertices)]
 
    # A utility function to check
    # if the current color assignment
    # is safe for vertex v
    def isSafe(self, v, colour, c):
        for i in range(self.V):
            if self.graph[v][i] == 1 and colour[i] == c:
                return False
        return True
     
    # A recursive utility function to solve m
    # coloring  problem
    def graphColourUtil(self, m, colour, v):
        if v == self.V:
            return True
 
        for c in range(1, m + 1):
            if self.isSafe(v, colour, c) == True:
                colour[v] = c
                if self.graphColourUtil(m, colour, v + 1) == True:
                    return True
                colour[v] = 0
 
    def graphColouring(self, m):
        colour = [0] * self.V
        if self.graphColourUtil(m, colour, 0) == None:
            return False
 
        # Print the solution
        print "Solution exist and Following
                  are the assigned colours:"
        for c in colour:
            print c,
        return True
 
# Driver Code
g = Graph(4)
g.graph = [[0, 1, 1, 1], [1, 0, 1, 0], [1, 1, 0, 1], [1, 0, 1, 0]]
m = 3
g.graphColouring(m)
 
# This code is contributed by Divyanshu Mehta

C＃

/* C# program for solution of M Coloring problem
using backtracking */
using System;
 
class GFG {
    readonly int V = 4;
    int[] color;
 
    /* A utility function to check if the current
    color assignment is safe for vertex v */
    bool isSafe(int v, int[, ] graph,
                int[] color, int c)
    {
        for (int i = 0; i < V; i++)
            if (graph[v, i] == 1 && c == color[i])
                return false;
        return true;
    }
 
    /* A recursive utility function to solve m
    coloring problem */
    bool graphColoringUtil(int[, ] graph, int m,
                           int[] color, int v)
    {
        /* base case: If all vertices are assigned
        a color then return true */
        if (v == V)
            return true;
 
        /* Consider this vertex v and try different
        colors */
        for (int c = 1; c <= m; c++) {
            /* Check if assignment of color c to v
            is fine*/
            if (isSafe(v, graph, color, c)) {
                color[v] = c;
 
                /* recur to assign colors to rest
                of the vertices */
                if (graphColoringUtil(graph, m,
                                      color, v + 1))
                    return true;
 
                /* If assigning color c doesn't lead
                to a solution then remove it */
                color[v] = 0;
            }
        }
 
        /* If no color can be assigned to this vertex
        then return false */
        return false;
    }
 
    /* This function solves the m Coloring problem using
    Backtracking. It mainly uses graphColoringUtil()
    to solve the problem. It returns false if the m
    colors cannot be assigned, otherwise return true
    and prints assignments of colors to all vertices.
    Please note that there may be more than one
    solutions, this function prints one of the
    feasible solutions.*/
    bool graphColoring(int[, ] graph, int m)
    {
        // Initialize all color values as 0. This
        // initialization is needed correct functioning
        // of isSafe()
        color = new int[V];
        for (int i = 0; i < V; i++)
            color[i] = 0;
 
        // Call graphColoringUtil() for vertex 0
        if (!graphColoringUtil(graph, m, color, 0)) {
            Console.WriteLine("Solution does not exist");
            return false;
        }
 
        // Print the solution
        printSolution(color);
        return true;
    }
 
    /* A utility function to print solution */
    void printSolution(int[] color)
    {
        Console.WriteLine("Solution Exists: Following"
                          + " are the assigned colors");
        for (int i = 0; i < V; i++)
            Console.Write(" " + color[i] + " ");
        Console.WriteLine();
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        GFG Coloring = new GFG();
 
        /* Create following graph and test whether it is
        3 colorable
        (3)---(2)
        | / |
        | / |
        | / |
        (0)---(1)
        */
        int[, ] graph = { { 0, 1, 1, 1 },
                          { 1, 0, 1, 0 },
                          { 1, 1, 0, 1 },
                          { 1, 0, 1, 0 } };
        int m = 3; // Number of colors
        Coloring.graphColoring(graph, m);
    }
}
 
// This code is contributed by PrinciRaj1992

输出

Solution Exists: Following are the assigned colors 
 1  2  3  2 

复杂度分析：

• 时间复杂度： O(m ^ V)。
  总共有O(m ^ V)种颜色组合。因此，时间复杂度为O(m ^ V)。上限时间复杂度保持不变，但平均花费的时间会更少。
• 空间复杂度： O(V)。
  graphColoring(…)函数的递归堆栈将需要O(V)空间。

方法3：使用BFS

此处的方法是首先用颜色1为每个节点从1着色为n。然后从未访问的起始节点开始行进BFS，以一次性覆盖所有连接的组件。在BFS遍历期间到达每个节点时，请执行以下操作：

• 检查给定节点的所有边缘。
• 对于通过一条边连接到我们节点的每个顶点：
  • 检查节点的颜色是否相同。如果相同，则将另一个节点(不是当前节点)的颜色增加一个。
  • 检查它是否访问过或未访问过。如果未访问，则将其标记为已访问并将其推入队列。
• 直到现在检查maxColors的条件。如果超过M，则返回false

访问所有节点后，返回true(因为在旅行时找不到违反条件)。

C++

// CPP program for the above approach
#include 
#include 
using namespace std;
 
class node
{
 
    // A node class which stores the color and the edges
    // connected to the node
public:
    int color = 1;
    set edges;
};
 
int canPaint(vector& nodes, int n, int m)
{
 
    // Create a visited array of n
    // nodes, initialized to zero
    vector visited(n + 1, 0);
 
    // maxColors used till now are 1 as
    // all nodes are painted color 1
    int maxColors = 1;
 
    // Do a full BFS traversal from
    // all unvisited starting points
    for (int sv = 1; sv <= n; sv++)
    {
 
        if (visited[sv])
            continue;
 
        // If the starting point is unvisited,
        // mark it visited and push it in queue
        visited[sv] = 1;
        queue q;
        q.push(sv);
 
        // BFS Travel starts here
        while (!q.empty())
        {
 
            int top = q.front();
            q.pop();
 
            // Checking all adjacent nodes
            // to "top" edge in our queue
            for (auto it = nodes[top].edges.begin();
                 it != nodes[top].edges.end(); it++)
            {
 
                // IMPORTANT: If the color of the
                // adjacent node is same, increase it by 1
                if (nodes[top].color == nodes[*it].color)
                    nodes[*it].color += 1;
 
                // If number of colors used shoots m, return
                // 0
                maxColors
                    = max(maxColors, max(nodes[top].color,
                                         nodes[*it].color));
                if (maxColors > m)
                    return 0;
 
                // If the adjacent node is not visited,
                // mark it visited and push it in queue
                if (!visited[*it]) {
                    visited[*it] = 1;
                    q.push(*it);
                }
            }
        }
    }
 
    return 1;
}
 
// Driver code
int main()
{
     
     int n = 4;
     bool graph[n][n] = {
     { 0, 1, 1, 1 },
     { 1, 0, 1, 0 },
     { 1, 1, 0, 1 },
     { 1, 0, 1, 0 }};
     int m = 3; // Number of colors
 
         
      // Create a vector of n+1
      // nodes of type "node"
      // The zeroth position is just
      // dummy (1 to n to be used)
      vector nodes(n + 1);
 
      // Add edges to each node as per given input
      for (int i = 0; i < n; i++)
      {
         for(int j =0;j

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class Node
{
    
    // A node class which stores the color and the edges
    // connected to the node
    int color = 1;
    Set edges = new HashSet();
}
class GFG
{
    static int canPaint(ArrayList nodes, int n, int m)
    {
       
        // Create a visited array of n
        // nodes, initialized to zero
        ArrayList visited = new ArrayList();
        for(int i = 0; i < n + 1; i++)
        {
            visited.add(0);
        }
         
        // maxColors used till now are 1 as
        // all nodes are painted color 1
        int maxColors = 1;
         
        // Do a full BFS traversal from
        // all unvisited starting points
        for (int sv = 1; sv <= n; sv++)
        {
            if (visited.get(sv) > 0)
            {
                continue;
            }
           
            // If the starting point is unvisited,
            // mark it visited and push it in queue
            visited.set(sv, 1);
            Queue q = new LinkedList<>();
            q.add(sv);
             
            // BFS Travel starts here
            while(q.size() != 0)
            {
                int top = q.peek();
                q.remove();
                 
                // Checking all adjacent nodes
                // to "top" edge in our queue
                for(int it: nodes.get(top).edges)
                {
                   
                    // IMPORTANT: If the color of the
                    // adjacent node is same, increase it by 1
                    if(nodes.get(top).color == nodes.get(it).color)
                    {
                        nodes.get(it).color += 1;
                    }
                   
                    // If number of colors used shoots m, return
                    // 0
                    maxColors = Math.max(maxColors,
                                        Math.max(nodes.get(top).color,
                                                 nodes.get(it).color));
                    if (maxColors > m)
                        return 0;
                   
                    // If the adjacent node is not visited,
                    // mark it visited and push it in queue
                    if (visited.get(it) == 0)
                    {
                        visited.set(it, 1);
                        q.remove(it);
                    }
                }
                 
            }
        }
        return 1;
    }
   
    // Driver code
    public static void main (String[] args)
    {
        int n = 4;
        int [][] graph = {{ 0, 1, 1, 1 },{ 1, 0, 1, 0 },
                          { 1, 1, 0, 1 },{ 1, 0, 1, 0 }};
        int m = 3; // Number of colors
         
        // Create a vector of n+1
        // nodes of type "node"
        // The zeroth position is just
        // dummy (1 to n to be used)
        ArrayList nodes = new ArrayList();
         
        for(int i = 0; i < n+ 1; i++)
        {
            nodes.add(new Node());
        }
         
        // Add edges to each node as per given input
      for (int i = 0; i < n; i++)
      {
         for(int j = 0; j < n; j++)
         {
             if(graph[i][j] > 0)
             {
                 // Connect the undirected graph
                  nodes.get(i).edges.add(i);
                  nodes.get(j).edges.add(j);
             }
         }
      }
       
      // Display final answer
        System.out.println(canPaint(nodes, n, m));
    }
}
 
// This code is contributed by avanitrachhadiya2155

Python3

# Python3 program for the above approach
from queue import Queue
 
class node:
     
    color = 1
    edges = set()
 
def canPaint(nodes, n, m):
 
    # Create a visited array of n
    # nodes, initialized to zero
    visited = [0 for _ in range(n+1)]
 
    # maxColors used till now are 1 as
    # all nodes are painted color 1
    maxColors = 1
 
    # Do a full BFS traversal from
    # all unvisited starting points
    for _ in range(1, n + 1):
        if visited[_]:
            continue
 
        # If the starting point is unvisited,
        # mark it visited and push it in queue
        visited[_] = 1
        q = Queue()
        q.put(_)
 
        # BFS Travel starts here
        while not q.empty():
            top = q.get()
 
            # Checking all adjacent nodes
            # to "top" edge in our queue
            for _ in nodes[top].edges:
 
                # IMPORTANT: If the color of the
                # adjacent node is same, increase it by 1
 
                if nodes[top].color == nodes[_].color:
                    nodes[_].color += 1
 
                # If number of colors used shoots m,
                # return 0
                maxColors = max(maxColors, max(
                    nodes[top].color, nodes[_].color))
                     
                if maxColors > m:
                    print(maxColors)
                    return 0
 
                # If the adjacent node is not visited,
                # mark it visited and push it in queue
                if not visited[_]:
                    visited[_] = 1
                    q.put(_)
                     
    return 1
 
# Driver code
if __name__ == "__main__":
     
    n = 4
    graph = [ [ 0, 1, 1, 1 ],
              [ 1, 0, 1, 0 ],
              [ 1, 1, 0, 1 ],
              [ 1, 0, 1, 0 ] ]
               
    # Number of colors
    m = 3 
 
    # Create a vector of n+1
    # nodes of type "node"
    # The zeroth position is just
    # dummy (1 to n to be used)
    nodes = []
    for _ in range(n+1):
        nodes.append(node())
 
    # Add edges to each node as
    # per given input
    for _ in range(n):
        for __ in range(n):
            if graph[_][__]:
                 
                # Connect the undirected graph
                nodes[_].edges.add(_)
                nodes[__].edges.add(__)
 
    # Display final answer
    print(canPaint(nodes, n, m))
 
# This code is contributed by harshitkap00r

C＃

// C# program for the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
 
class node
{
   
    // A node class which stores the color and the edges
    // connected to the node
    public int color = 1;
    public HashSet edges = new HashSet();
};
 
static int canPaint(List nodes, int n, int m)
{
 
    // Create a visited array of n
    // nodes, initialized to zero
    List visited = new List();
 
    for(int i = 0; i < n + 1; i++)
    {
        visited.Add(0);
    }
 
    // maxColors used till now are 1 as
    // all nodes are painted color 1
    int maxColors = 1;
 
    // Do a full BFS traversal from
    // all unvisited starting points
    for (int sv = 1; sv <= n; sv++)
    {
 
        if (visited[sv] > 0)
            continue;
 
        // If the starting point is unvisited,
        // mark it visited and push it in queue
        visited[sv] = 1;
        Queue q = new Queue();
        q.Enqueue(sv);
 
        // BFS Travel starts here
        while (q.Count != 0)
        {
 
            int top = (int)q.Peek();
            q.Dequeue();
 
            // Checking all adjacent nodes
            // to "top" edge in our queue
            foreach(int it in nodes[top].edges)
            {
               
                // IMPORTANT: If the color of the
                // adjacent node is same, increase it by 1
                if (nodes[top].color == nodes[it].color)
                    nodes[it].color += 1;
 
                // If number of colors used shoots m, return
                // 0
                maxColors
                    = Math.Max(maxColors, Math.Max(nodes[top].color,
                                         nodes[it].color));
                if (maxColors > m)
                    return 0;
 
                // If the adjacent node is not visited,
                // mark it visited and push it in queue
                if (visited[it] == 0)
                {
                    visited[it] = 1;
                    q.Enqueue(it);
                }
            }
        }
    }
    return 1;
}
 
// Driver code
public static void Main()
{
     
     int n = 4;
     int [,]graph = {
     { 0, 1, 1, 1 },
     { 1, 0, 1, 0 },
     { 1, 1, 0, 1 },
     { 1, 0, 1, 0 }};
     int m = 3; // Number of colors
     
      // Create a vector of n+1
      // nodes of type "node"
      // The zeroth position is just
      // dummy (1 to n to be used)
      List nodes = new List();
      for(int i = 0; i < n+ 1; i++)
      {
        nodes.Add(new node());
      }
 
 
      // Add edges to each node as per given input
      for (int i = 0; i < n; i++)
      {
         for(int j = 0; j < n; j++)
         {
             if(graph[i, j] > 0)
             {
                
                  // Connect the undirected graph
                  nodes[i].edges.Add(i);
                  nodes[j].edges.Add(j);
              }
         }
      }
 
        // Display final answer
        Console.WriteLine(canPaint(nodes, n, m));
   }
}
 
 
// Thiis code is contributed by rutvik_56.

输出

1

复杂度分析：

• 时间复杂度： O(V + E)。
• 空间复杂度： O(V)。用于存储访问列表。