HDU 2120--Ice_cream's world I【并查集， 判断环的个数】

Ice_cream's world I

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

Total Submission(s): 783 Accepted Submission(s): 458



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 <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;
#define maxn 1100
int N, sum, M;
int per[maxn];

void init() {
    sum = 0;
    for(int i = 0; i < N; ++i)
        per[i] = i;
}

int find(int x){
    int r = x;
    while( r != per[r])
        r = per[r];
    per[x] = r;
    return r;
}

void join (int a, int b){
    int fa = find(a);
    int fb = find(b);
    if(fa != fb){
        per[fa] = fb;
    }
    else sum++;
}

int main (){
    while(scanf("%d%d", &N, &M) != EOF){
        init();
        while(M--){
            int a, b;
            scanf("%d%d", &a, &b);
            join(a, b);
        }
        printf("%d\n", sum);
    }
    return 0;
}