Graph’s Cycle Component

最新推荐文章于 2025-06-03 21:08:22 发布

Gabriel_tzc

最新推荐文章于 2025-06-03 21:08:22 发布

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本文介绍了一种算法，用于计算给定图中各组件的数量及其中环形图的数量。通过输入顶点数量和边的关系，算法能够找出直接或间接相连的顶点所构成的组件，并判断哪些组件构成了环形图。

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描述：

In graph theory, a cycle graph is an undirected graph that consists of a single cycle, or in other words, some number of vertices connected in a closed chain.
Now, you are given a graph where some vertices are connected to be components, can you figure out how many components are there in the graph and how many of those components are cycle graphs.
Two vertices belong to a same component if and only if those two vertices connect each other directly or indirectly.

输出：

The input consists of multiply test cases.
The first line of each test case contains two integer, n (0 < n < 100000), m (0 <= m <= 300000), which are the number of vertices and the number of edges.
The next m lines, each line consists of two integers, u, v, which means there is an edge between u and v.
You can assume that there is no multiply edges and no loops.
The last test case is followed by two zeros, which means the end of input.

输出：

For each test case, output the number of all the components and the number of components which are cycle graphs.

样例输出：

8 9
0 1
1 3
2 3
0 2
4 5
5 7
6 7
4 6
4 7
2 1
0 1
0 0

样例输出：

2 1
1 0

题目大意：

您将获得一些图形，其中某些顶点连接到组件，您可以确定图中有多少组件，以及这些组件中有多少是循环图。当且仅当这两个顶点直接或间接连接时，两个顶点属于相同的组件。

#include<stdio.h>
#include<string.h>
int map[110000],asd[110000];
int n,m;
int find(int x)											//查找祖先结点 
{
    return x==map[x]?x:find(map[x]);
}
void combil(int a,int b)								//建立连接使得有共同祖先 
{
    int p=find(a);
    int q=find(b);
	if(p!=q)
	{
		if(p<q)
        map[q]=p;
   		else
        map[p]=q;
	}
}
int main()
{
	int a,b,cycle,conponents;
    while(scanf("%d %d",&n,&m),n)
    {
        for(int i=1;i<=n;i++)
        map[i]=i;
        memset(asd,0,sizeof(asd));
        for(int i=1;i<=m;i++)
        {
            scanf("%d %d",&a,&b);
            asd[++a]++;											//注意要先加加 
            asd[++b]++;
            combil(a,b);
        }
        conponents=0;
        cycle=0;
        for(int i=1;i<=n;i++)
        {
	     if(map[i]==i)
             conponents++;
	}
        for(int i=1;i<=n;i++)
        {
            if(asd[i]!=2)										//如果个数不是2就一定不是组件 
            map[find(i)]=0;
        }
        for(int i=1;i<=n;i++)
	{
            if(map[i]==i)
            cycle++;			
	}
        printf("%d %d\n",conponents,cycle);
    }
    return 0;
}