其实就是判断新加的点是否在同一个集合里面

代码虐我千百遍，我待代码入初恋

Ice_cream‘s world I

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

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>
#include <stdlib.h>
#include <malloc.h>
#include <limits.h>
#include <ctype.h>
#include <string.h>
#include <string>
#include <queue>
#include <algorithm>
#include <iostream>
#include <stack>
#include <deque>
#include <vector>
#include <set>
#include <map>
using namespace std;
#define MAXN 1000 + 10
#define MAXN1 10000 + 10
int father[MAXN];
int a[MAXN1];
int b[MAXN1];

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

    return father[x];
}

int main(){
    int n,m;
    int i;

    while(~scanf("%d%d",&n,&m)){
        for(i=0;i<MAXN;i++){
            father[i] = i;
        }
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=0;i<m;i++){
            scanf("%d%d",&a[i],&b[i]);
        }
        int sum = 0;
        for(i=0;i<m;i++){
            int f1 = find(a[i]);
            int f2 = find(b[i]);
            if(f1 != f2){
                if(f1 < f2){
                    father[f2] = f1;
                }
                else{
                    father[f1] = f2;
                }
            }
            else{
                sum++;
            }
        }
        printf("%d\n",sum);
    }

    return 0;
}