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): 355    Accepted Submission(s): 181
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
 
 
Source
HDU 2007-10 Programming Contest_WarmUp
 
 
Recommend
ウイスキー
 
タイトルがよく見えなかったので、、、GOD
 
分割された領域の数を検索するためにセットの適用を調べます.
すなわち、2つのノードの値が同じである場合、1つのリングのためであることを示すが、そうでない場合、領域の個数に1を加算することは不可能である.
 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int maxn=1010;

int n,m;
int father[maxn];

void makeSet(){
    for(int i=0;i<n;i++){
        father[i]=i;
    }
}

int findSet(int x){
    if(x!=father[x]){
        father[x]=findSet(father[x]);
    }
    return father[x];
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d",&n,&m)){
        makeSet();
        int ans=0;
        int a,b;
        while(m--){
            scanf("%d%d",&a,&b);
            int fx=findSet(a);
            int fy=findSet(b);
            if(fx==fy)
                ans++;
            else
                father[fx]=fy;
        }
        printf("%d
",ans);
    }
    return 0;
}