Hdu 2120 Ice_cream's world I

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1611    Accepted Submission(s): 953

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

Output

Output the maximum number of ACMers who will be awarded.
One answer one line.

Sample Input

8 10

0 1

1 2

1 3

2 4

3 4

0 5

5 6

6 7

3 6

4 7

Sample Output

3

#include <stdio.h>
int fa[1000+10];
int n;
void init()
{
    for(int i=0;i<=n;i++)
    fa[i]=i;
}
int find(int x)
{
    if(fa[x]!=x) fa[x]=find(fa[x]);
    return fa[x];
}
int main()
{
    int m;
    while(~scanf("%d %d",&n,&m))
    {
        int result=0;
        init();
        for(int i=0;i<m;i++)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            int x=find(a);
            int y=find(b);
            if(x!=y)
            {
                fa[x]=y;
            }
            else//成环，但是不将这条边加到集合中去，记录成环一次             result++;
        }
        printf("%d
",result);
    }
    return 0;
}