HDU 3560 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): 1655    Accepted Submission(s): 637

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

 

 

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思路：并查集+环路检查，我竟然不知道what is a cycle graph?看了别人的说法才知道是一个什么东西，然后我觉得应该先求出连通分量，然后求cycle graph，

但是竟然不行，只能使用并查集，但是并查集怎么可能保证这样呢，Two vertices belong to a same component if and only if those two vertices connect

each other directly or indirectly，并查集确定的component，不能保证connect each other 吧。

 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
int map[110000];
int indegree[110000];
int the_last_conponent;
int the_last_cycle;
int n,m;
int a,b;
int find(int x)
{
    while(x != map[x])
    {
        x = map[x];
    }
    return x;
}
void Merge(int a,int b)
{
    int p = find(a);
    int q = find(b);
    if(p < q)
        map[q] = p;
    else
        map[p] = q;
}
int main()
{
    while(scanf("%d%d",&n,&m) != EOF && (n + m))
    {
        for(int i = 1;i <= n;i ++)
           map[i] = i;
        memset(indegree,0,sizeof(indegree));
        for(int i = 1;i <= m;i ++)
        {
            scanf("%d%d",&a,&b);
            a ++;b ++;
            indegree[a] ++;
            indegree[b] ++;
            Merge(a,b);
        }
        the_last_conponent = 0;
        the_last_cycle = 0;
        for(int i = 1;i <= n;i ++)
            if(map[i] == i)
                the_last_conponent ++;
        for(int i = 1; i <= n;i ++)
        {
            if(indegree[i] != 2)
                map[find(i)] = 0;
        }
        for(int i = 1;i <= n;i ++)
            if(map[i] == i)
                the_last_cycle ++;
        printf("%d %d\n",the_last_conponent,the_last_cycle);
    }
    return 0;
}