hdoj 2120 Ice_cream's world I　【并查集判断成环数】

Ice_cream's world I

[b]Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 701 Accepted Submission(s): 401

[/b]

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

题意：判断成环数。

并查集实现代码：

#include<stdio.h>
#include<string.h>
#define max 1000+10
int set[max];
int sum=0;
int find(int p)
{
    int child=p;
    int t;
    while(p!=set[p])
    p=set[p];
    while(child!=p)
    {
        t=set[child];
        set[child]=p;
        child=t;
    }
    return p;
}
void merge(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    set[fx]=fy;
    else
    sum++;
}
int main()
{
    int n,m,i,j,x,y;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<n;i++)
        set[i]=i;
        while(m--)
        {
            scanf("%d%d",&x,&y);
            merge(x,y);
        }
        printf("%d\n",sum);
        sum=0;
    }
    return 0;
}