Graph’s Cycle Component

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Problem Description

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.

Input

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.

Output

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

Sample Input

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

Sample Output

2 1
1 0

Author

[email protected]

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host
by WHU

Recommend

zhouzeyong   |   We have carefully selected several similar problems for you:  3559 3558 3563 3562 3561

ac代码

#include<stdio.h>
#include<string.h>
int pre[100010],dig[100010];
int n,m;
/*int find(int x)
{
	int r=x;
	while(r!=pre[r])
		r=pre[r];
	int j,k;
	j=x;
	while(j!=r)
	{
		k=pre[j];
		pre[j]=r;
		j=k;
	}
	return r;
}*/
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF,n||m)
	{
		//init();
		//memset(dig,0,sizeof(dig));
		int i,j;
		for(i=1;i<=n;i++)
		{
			pre[i]=i;
			dig[i]=0;
		}
		for(i=0;i<m;i++)
		{
			int u,v;
			scanf("%d%d",&u,&v);
			u++;v++;
			dig[u]++;
			dig[v]++;
			//memge(u,v);
			int fa=u;
			while(fa!=pre[fa])
				fa=pre[fa];
			int fb=v;
			while(fb!=pre[fb])
				fb=pre[fb];
			if(fa!=fb)
			{
				if(fa<fb)
				{
					pre[fb]=fa;
				}
				else
					pre[fa]=fb;
			}
		}
		int ans1=0,ans2=0;
		for(i=1;i<=n;i++)
		{
			if(pre[i]==i)
			{
				ans1++;
			}
		}
		for(i=1;i<=n;i++)
		{
			if(dig[i]!=2)
			{
				int fi=i;
				while(fi!=pre[fi])
				{
					fi=pre[fi];
				}
				pre[fi]=0;//设为-1会超时，，，，，
			}
		}
		for(i=1;i<=n;i++)
		{
			if(pre[i]==i)
				ans2++;
		}
		printf("%d %d\n",ans1,ans2);
	}
}