📜  所需的最少颜色,以便边缘形成周期没有相同的颜色

📅  最后修改于: 2021-10-25 05:01:47             🧑  作者: Mango

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:

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_56


C#
// 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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