HDU2120Ice_cream's world I（基础并查集）

Ice_cream's world I

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

Total Submission(s): 698 Accepted Submission(s): 398



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

Author
Wiskey
题意：就是图中有几个环。

#include<stdio.h>
const int N = 1015;
int fath
,n;

void init()
{
    for(int i =0;i<=n;i++)
        fath[i]=i;
}
int findfath(int x)
{
    if(x==fath[x])
        return fath[x];
    fath[x]=findfath(fath[x]);
    return fath[x];
}
int setfath(int x,int y)
{
    x=findfath(x);
    y=findfath(y);
    if(x==y)
        return 1;
    fath[x]= y;
    return 0;
}
int main()
{
    int x,y,m,ans;
    while(scanf("%d%d",&n,&m)>0)
    {
        init();
        ans=0;
        while(m--)
        {
            scanf("%d%d",&x,&y);
            ans+=setfath(x,y);
        }
        printf("%d\n",ans);
    }

}