HDU2120【并查集判环】

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 600    Accepted Submission(s): 344

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

 

大意：

题意很简单，刚开始我理解成两个塔之间有一个墙把他们隔开呢T_T

 

分析：并查集判环  只要在一个集合之中一定会有一块空地

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 1005;

int fa[maxn];

int find(int x) {
    if(fa[x] == x) return x;
    return fa[x] = find(fa[x]);
}

int main() {
    int n, m;
    int a, b;
    while(EOF != scanf("%d %d",&n, &m) ) {
        for(int i = 0; i < n; i++) fa[i] = i;
        int cnt = 0;
        while(m--) {
            scanf("%d %d",&a, &b);
            if(find(a) == find(b) ) {
                cnt++;
            } else {
                fa[find(a)] = find(b);
            }
        }
        printf("%d\n", cnt);
    }
    return 0;
}