给定一个有V个顶点和E 条边的有向图，没有自环和多条边，任务是找到所需的最小颜色数，使得相同颜色的边不形成循环，并找到每条边的颜色。
例子：
Input: V = {1, 2, 3}, E = {(1, 2), (2, 3), (3, 1)}
Output: 2
1 1 2
Explanation:
In the above given graph it forms only one cycle,
that is Vertices connecting 1, 2, 3 forms a cycle
Then the edges connecting 1->2 or 2->3 or 3->1 can be colored
with a different color such that edges forming cycle don’t have same color
Input: V = {1, 2, 3, 4, 5}, E = {(1, 2), (1, 3), (2, 3), (2, 4), (3, 4), (4, 5), (5, 3)}
Output: 2
Colors of Edges – 1 1 1 1 1 1 2
Explanation:
In the above given graph it forms only one cycle,
that is Vertices connecting 3, 4, 5 forms a cycle
Then the edges connecting 5->3 or 4->5 or 3->4 can be colored
with a different color such that edges forming cycle don’t have same color
Final Colors of the Edges –
{1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 2}
The above array denotes the pairs as – Edge : Color Code
方法：这个想法是在图中找到循环，这可以在 DFS 的帮助下完成，其中当一个已经访问过的节点用新边再次访问时，该边用另一种颜色着色否则，如果没有循环，则所有边缘都只能用一种颜色着色。
算法：
- 用颜色 1 标记每条边，并将每个顶点标记为未访问。
- 使用图的 DFS Traversal 遍历图并标记访问过的节点。
- 当一个节点已经被访问过时，连接顶点的边被标记为颜色 2。
- 访问所有顶点时打印边的颜色。
举例说明：
示例 1 的详细试运行
| Current Vertex | Current Edge | Visited Vertices | Colors of Edges | Comments |
|---|---|---|---|---|
| 1 | 1–>2 | {1} | {1: 1, 2: 1, 3: 1} | Node 1 is marked as visited and Calling DFS for node 2 |
| 2 | 2–>3 | {1, 2} | {1: 1, 2: 1, 3: 1} | Node 2 is marked as visited and Calling DFS for node 3 |
| 3 | 3–>1 | {1, 2} | {1: 1, 2: 1, 3: 2} | As 1 is already Visited color of Edge 3 is changed to 2 |
下面是上述方法的实现：
C++
// C++ implementation to find the
// minimum colors required to
// such that edges forming cycle
// don't have same color
#include
using namespace std;
const int n = 5, m = 7;
// Variable to store the graph
vector > g[m];
// To store that the
// vertex is visited or not
int col[n];
// Boolean Value to store that
// graph contains cycle or not
bool cyc;
// Variable to store the color
// of the edges of the graph
int res[m];
// Function to traverse graph
// using DFS Traversal
void dfs(int v)
{
col[v] = 1;
// Loop to iterate for all
// edges from the source vertex
for (auto p : g[v]) {
int to = p.first, id = p.second;
// If the vertex is not visited
if (col[to] == 0)
{
dfs(to);
res[id] = 1;
}
// Condition to check cross and
// forward edges of the graph
else if (col[to] == 2)
{
res[id] = 1;
}
// Presence of Back Edge
else {
res[id] = 2;
cyc = true;
}
}
col[v] = 2;
}
// Driver Code
int main()
{
g[0].push_back(make_pair(1, 0));
g[0].push_back(make_pair(2, 1));
g[1].push_back(make_pair(2, 2));
g[1].push_back(make_pair(3, 3));
g[2].push_back(make_pair(3, 4));
g[3].push_back(make_pair(4, 5));
g[4].push_back(make_pair(2, 6));
// Loop to run DFS Traversal on
// vertex which is not visited
for (int i = 0; i < n; ++i) {
if (col[i] == 0)
{
dfs(i);
}
}
cout << (cyc ? 2 : 1) << endl;
// Loop to print the
// colors of the edges
for (int i = 0; i < m; ++i) {
cout << res[i] << ' ';
}
return 0;
} Java
// Java implementation to find the
// minimum colors required to
// such that edges forming cycle
// don't have same color
import java.util.*;
class GFG{
static int n = 5, m = 7;
static class pair
{
int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Variable to store the graph
static Vector []g = new Vector[m];
// To store that the
// vertex is visited or not
static int []col = new int[n];
// Boolean Value to store that
// graph contains cycle or not
static boolean cyc;
// Variable to store the color
// of the edges of the graph
static int []res = new int[m];
// Function to traverse graph
// using DFS Traversal
static void dfs(int v)
{
col[v] = 1;
// Loop to iterate for all
// edges from the source vertex
for (pair p : g[v]) {
int to = p.first, id = p.second;
// If the vertex is not visited
if (col[to] == 0)
{
dfs(to);
res[id] = 1;
}
// Condition to check cross and
// forward edges of the graph
else if (col[to] == 2)
{
res[id] = 1;
}
// Presence of Back Edge
else {
res[id] = 2;
cyc = true;
}
}
col[v] = 2;
}
// Driver Code
public static void main(String[] args)
{
for(int i= 0; i < m; i++)
g[i] = new Vector();
g[0].add(new pair(1, 0));
g[0].add(new pair(2, 1));
g[1].add(new pair(2, 2));
g[1].add(new pair(3, 3));
g[2].add(new pair(3, 4));
g[3].add(new pair(4, 5));
g[4].add(new pair(2, 6));
// Loop to run DFS Traversal on
// vertex which is not visited
for (int i = 0; i < n; ++i) {
if (col[i] == 0)
{
dfs(i);
}
}
System.out.print((cyc ? 2 : 1) +"\n");
// Loop to print the
// colors of the edges
for (int i = 0; i < m; ++i) {
System.out.print(res[i] +" ");
}
}
}
// This code is contributed by sapnasingh4991 Python3
# Python3 implementation to find the
# minimum colors required to
# such that edges forming cycle
# don't have same color
n = 5
m = 7;
# Variable to store the graph
g = [[] for i in range(m)]
# To store that the
# vertex is visited or not
col = [0 for i in range(n)];
# Boolean Value to store that
# graph contains cycle or not
cyc = True
# Variable to store the color
# of the edges of the graph
res = [0 for i in range(m)];
# Function to traverse graph
# using DFS Traversal
def dfs(v):
col[v] = 1;
# Loop to iterate for all
# edges from the source vertex
for p in g[v]:
to = p[0]
id = p[1];
# If the vertex is not visited
if (col[to] == 0):
dfs(to);
res[id] = 1;
# Condition to check cross and
# forward edges of the graph
elif (col[to] == 2):
res[id] = 1;
# Presence of Back Edge
else:
res[id] = 2;
cyc = True;
col[v] = 2;
# Driver Code
if __name__=='__main__':
g[0].append([1, 0]);
g[0].append([2, 1]);
g[1].append([2, 2]);
g[1].append([3, 3]);
g[2].append([3, 4]);
g[3].append([4, 5]);
g[4].append([2, 6]);
# Loop to run DFS Traversal on
# vertex which is not visited
for i in range(n):
if (col[i] == 0):
dfs(i);
print(2 if cyc else 1)
# Loop to print the
# colors of the edges
for i in range(m):
print(res[i], end=' ')
# This code is contributed by rutvik_56C#
// C# implementation to find the
// minimum colors required to
// such that edges forming cycle
// don't have same color
using System;
using System.Collections.Generic;
class GFG{
static int n = 5, m = 7;
class pair
{
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
// Variable to store the graph
static List []g = new List[m];
// To store that the
// vertex is visited or not
static int []col = new int[n];
// Boolean Value to store that
// graph contains cycle or not
static bool cyc;
// Variable to store the color
// of the edges of the graph
static int []res = new int[m];
// Function to traverse graph
// using DFS Traversal
static void dfs(int v)
{
col[v] = 1;
// Loop to iterate for all
// edges from the source vertex
foreach (pair p in g[v]) {
int to = p.first, id = p.second;
// If the vertex is not visited
if (col[to] == 0)
{
dfs(to);
res[id] = 1;
}
// Condition to check cross and
// forward edges of the graph
else if (col[to] == 2)
{
res[id] = 1;
}
// Presence of Back Edge
else {
res[id] = 2;
cyc = true;
}
}
col[v] = 2;
}
// Driver Code
public static void Main(String[] args)
{
for(int i= 0; i < m; i++)
g[i] = new List();
g[0].Add(new pair(1, 0));
g[0].Add(new pair(2, 1));
g[1].Add(new pair(2, 2));
g[1].Add(new pair(3, 3));
g[2].Add(new pair(3, 4));
g[3].Add(new pair(4, 5));
g[4].Add(new pair(2, 6));
// Loop to run DFS Traversal on
// vertex which is not visited
for (int i = 0; i < n; ++i) {
if (col[i] == 0)
{
dfs(i);
}
}
Console.Write((cyc ? 2 : 1) +"\n");
// Loop to print the
// colors of the edges
for (int i = 0; i < m; ++i) {
Console.Write(res[i] +" ");
}
}
}
// This code is contributed by PrinciRaj1992 Javascript
输出：
2
1 1 1 1 1 1 2性能分析：
- 时间复杂度：与上述方法一样，图的 DFS 遍历需要 O(V + E) 时间，其中 V 表示顶点数，E 表示边数。因此时间复杂度将为O(V + E) 。
- 辅助空间： O(1)。
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