在无向图中计算从源到达目的地的总方式

给定一个无向图、一个源顶点‘s’和一个目标顶点‘d’ ，任务是计算从给定‘s’到‘d’的总路径。

例子

Input: s = 1, d = 4  

Output: 2 
Explanation: 
Below are the 2 paths from 1 to 4 
1 -> 3 -> 4 
1 -> 2 -> 3 -> 4

Input: s = 3, d = 9 

Output: 6 
Explanation: 
Below are the 6 paths from 3 to 9 
3 -> 2 -> 1 -> 7 -> 6 -> 5 -> 10 -> 9 
3 -> 2 -> 1 -> 7 -> 6 -> 10 -> 9 
3 -> 2 -> 1 -> 7 -> 8 -> 9 
3 -> 4 -> 2 -> 1 -> 7 -> 6 -> 5 -> 10 -> 9 
3 -> 4 -> 2 -> 1 -> 7 -> 6 -> 10 -> 9 
3 -> 4 -> 2 -> 1 -> 7 -> 8 -> 9  

方法：
这个想法是对给定的无向图进行深度优先遍历。

• 从源头开始遍历。
• 继续将访问过的顶点存储在一个名为“visited[]”的数组中。
• 如果我们到达目标顶点，则将计数增加“1”。
• 重要的是将visited[]中的当前顶点标记为visited，这样遍历就不会循环进行。

下面是上述方法的实现：

C++

// C++ program to count total number of
// ways to reach destination in a graph
#include 
using namespace std;
 
// Utility Function to count total ways
int countWays(int mtrx[][11], int vrtx,
              int i, int dest, bool visited[])
{
    // Base condition
    // When reach to the destination
    if (i == dest) {
 
        return 1;
    }
    int total = 0;
    for (int j = 0; j < vrtx; j++) {
        if (mtrx[i][j] == 1 && !visited[j]) {
 
            // Make vertex visited
            visited[j] = true;
 
            // Recursive function, for count ways
            total += countWays(mtrx, vrtx,
                               j, dest, visited);
 
            // Backtracking
            // Make vertex unvisited
            visited[j] = false;
        }
    }
 
    // Return total ways
    return total;
}
 
// Function to count total ways
// to reach destination
int totalWays(int mtrx[][11], int vrtx,
              int src, int dest)
{
    bool visited[vrtx];
 
    // Loop to make all vertex unvisited,
    // Initially
    for (int i = 0; i < vrtx; i++) {
        visited[i] = false;
    }
 
    // Make source visited
    visited[src] = true;
 
    return countWays(mtrx, vrtx, src, dest,
                     visited);
}
 
int main()
{
    int vrtx = 11;
    int mtrx[11][11] = {
        { 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0 },
        { 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0 },
        { 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0 },
        { 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0 },
        { 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0 }
    };
 
    int src = 3;
    int dest = 9;
 
    // Print total ways
    cout << totalWays(mtrx, vrtx, src - 1,
                      dest - 1);
 
    return 0;
}

Java

// Java program to count total number of
// ways to reach destination in a graph
class GFG{
  
// Utility Function to count total ways
static int countWays(int mtrx[][], int vrtx,
              int i, int dest, boolean visited[])
{
    // Base condition
    // When reach to the destination
    if (i == dest) {
  
        return 1;
    }
    int total = 0;
    for (int j = 0; j < vrtx; j++) {
        if (mtrx[i][j] == 1 && !visited[j]) {
  
            // Make vertex visited
            visited[j] = true;
  
            // Recursive function, for count ways
            total += countWays(mtrx, vrtx,
                               j, dest, visited);
  
            // Backtracking
            // Make vertex unvisited
            visited[j] = false;
        }
    }
  
    // Return total ways
    return total;
}
  
// Function to count total ways
// to reach destination
static int totalWays(int mtrx[][], int vrtx,
              int src, int dest)
{
    boolean []visited = new boolean[vrtx];
  
    // Loop to make all vertex unvisited,
    // Initially
    for (int i = 0; i < vrtx; i++) {
        visited[i] = false;
    }
  
    // Make source visited
    visited[src] = true;
  
    return countWays(mtrx, vrtx, src, dest,
                     visited);
}
  
public static void main(String[] args)
{
    int vrtx = 11;
    int mtrx[][] = {
        { 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0 },
        { 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0 },
        { 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0 },
        { 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0 },
        { 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0 }
    };
  
    int src = 3;
    int dest = 9;
  
    // Print total ways
    System.out.print(totalWays(mtrx, vrtx, src - 1,
                      dest - 1));
  
}
}
 
// This code contributed by Rajput-Ji

Python 3

# Python 3 program to count total number of
# ways to reach destination in a graph
 
# Utility Function to count total ways
def countWays(mtrx, vrtx, i, dest, visited):
     
    # Base condition
    # When reach to the destination
    if (i == dest):
        return 1
         
    total = 0
    for j in range(vrtx):
        if (mtrx[i][j] == 1 and not visited[j]):
 
            # Make vertex visited
            visited[j] = True;
 
            # Recursive function, for count ways
            total += countWays(mtrx, vrtx, j, dest, visited);
 
            # Backtracking
            # Make vertex unvisited
            visited[j] = False;
 
    # Return total ways
    return total
 
# Function to count total ways
# to reach destination
def totalWays(mtrx, vrtx, src, dest):
    visited = [False]*vrtx
 
    # Loop to make all vertex unvisited,
    # Initially
    for i in range(vrtx):
        visited[i] = False
 
    # Make source visited
    visited[src] = True;
 
    return countWays(mtrx, vrtx, src, dest,visited)
 
# Driver function
vrtx = 11
mtrx = [
        [0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0 ],
        [ 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0 ],
        [ 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0 ],
        [ 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0 ],
        [ 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0 ],
        [ 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0 ],
        [ 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0 ],
        [ 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0 ],
        [ 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0 ],
        [ 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0 ],
        [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0 ]
    ]
 
src = 3
dest = 9
 
# Print total ways
print(totalWays(mtrx, vrtx, src - 1,dest - 1))
 
# This code is contributed by atul kumar shrivastava

C#

// C# program to count total number of
// ways to reach destination in a graph
using System;
 
class GFG{
   
// Utility Function to count total ways
static int countWays(int[,] mtrx, int vrtx,
              int i, int dest, bool[] visited)
{
    // Base condition
    // When reach to the destination
    if (i == dest) {
   
        return 1;
    }
    int total = 0;
    for (int j = 0; j < vrtx; j++) {
        if (mtrx[i,j] == 1 && !visited[j]) {
   
            // Make vertex visited
            visited[j] = true;
   
            // Recursive function, for count ways
            total += countWays(mtrx, vrtx,
                               j, dest, visited);
   
            // Backtracking
            // Make vertex unvisited
            visited[j] = false;
        }
    }
   
    // Return total ways
    return total;
}
   
// Function to count total ways
// to reach destination
static int totalWays(int[,] mtrx, int vrtx,
              int src, int dest)
{
    bool[]visited = new bool[vrtx];
   
    // Loop to make all vertex unvisited,
    // Initially
    for (int i = 0; i < vrtx; i++) {
        visited[i] = false;
    }
   
    // Make source visited
    visited[src] = true;
   
    return countWays(mtrx, vrtx, src, dest,
                     visited);
}
   
public static void Main()
{
    int vrtx = 11;
    int[,] mtrx = {
        { 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0 },
        { 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0 },
        { 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0 },
        { 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0 },
        { 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0 },
        { 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0 },
        { 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0 }
    };
   
    int src = 3;
    int dest = 9;
   
    // Print total ways
    Console.Write(totalWays(mtrx, vrtx, src - 1,
                      dest - 1));
   
}
}

Javascript

// Javascript program to count total number of
// ways to reach destination in a graph
 
// Utility Function to count total ways
function countWays(mtrx,vrtx,i,dest,visited)
{
    // Base condition
    // When reach to the destination
    if (i == dest) {
    
        return 1;
    }
    let total = 0;
    for (let j = 0; j < vrtx; j++) {
        if (mtrx[i][j] == 1 && !visited[j]) {
    
            // Make vertex visited
            visited[j] = true;
    
            // Recursive function, for count ways
            total += countWays(mtrx, vrtx,
                               j, dest, visited);
    
            // Backtracking
            // Make vertex unvisited
            visited[j] = false;
        }
    }
    
    // Return total ways
    return total;
}
 
// Function to count total ways
// to reach destination
function totalWays(mtrx,vrtx,src,dest)
{
    let visited = new Array(vrtx);
    
    // Loop to make all vertex unvisited,
    // Initially
    for (let i = 0; i < vrtx; i++) {
        visited[i] = false;
    }
    
    // Make source visited
    visited[src] = true;
    
    return countWays(mtrx, vrtx, src, dest,
                     visited);
}
 
let vrtx = 11;
let mtrx=[
        [ 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0 ],
        [ 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0 ],
        [ 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0 ],
        [ 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0 ],
        [ 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0 ],
        [ 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0 ],
        [ 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0 ],
        [ 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0 ],
        [ 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0 ],
        [ 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0 ],
        [ 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0 ]
    ];
     
let src = 3;
let dest = 9;
// Print total ways
document.write(totalWays(mtrx, vrtx, src - 1,
                      dest - 1));
// This code is contributed by avanitrachhadiya2155

输出：

6

性能分析：

• 时间复杂度：在上述方法中，对于给定的顶点，我们检查所有顶点，因此时间复杂度为O(N*N) ，其中 N 不是顶点。
• 辅助空间复杂度：在上述方法中，我们使用大小为 N 的访问数组，其中 N 是顶点数，因此辅助空间复杂度为O(N) 。

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。