Hdu Graph’s Cycle Component

Graph’s Cycle Component

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1275    Accepted Submission(s): 499

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

momodi@WHU

 

 

Source

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

 

//并查集判断连通分量数，连通分块的每个度数为2代表为圈图

#include <iostream>
#include <algorithm>
#include <queue>
#include <cmath>
#define N 100003
using namespace std;
int f[N];
int r[N];
int find_r(int x)
{
   if(f[x]>=0)
   {
     return  f[x]=find_r(f[x]);
   }
   return x;
}
void union_set(int x,int y)
{
    x=find_r(x);
	y=find_r(y);
	if(x==y) return ;
	if(r[x]>r[y])
	   f[y]=x;
	 else 
		 if(r[x]==r[y])  
		 {
	         f[y]=x;
			 r[x]++;
		 }
		 else
			 f[x]=y;

}
int de[N];
int main()
{
	int i,n,m;
	int x,y;
	int r_n,r_m;
   while(scanf("%d%d",&n,&m),n||m)
   {
	   for(i=0;i<n;i++)  //开始这用memset()函数显示超时、、看来、、、
	   {
	      de[i]=r[i]=0;
		  f[i]=-1;
	   }
       for(i=0;i<m;i++)
	   {
		   scanf("%d%d",&x,&y);
	        union_set(x,y);
			de[x]++;de[y]++;
	   }
	   
	   for(i=0;i<n;i++)
		   if(de[i]!=2)
		   {
		       r_m=find_r(i);
			   r[r_m]=-1;
		   }
		   r_n=r_m=0;
         for(i=0;i<n;i++)
		   if(f[i]==-1)
		   {
		       r_n++;
			   if(r[i]!=-1)
				   r_m++;
		   }
		printf("%d %d\n",r_n,r_m);
   }
   return 0;
}