HDOJ-----2120并查集（水题）

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1270    Accepted Submission(s): 761

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

8 100 11 21 32 43 40 55 66 73 64 7

 

Sample Output

3

看英文题就是头疼，看懂了原来就是个这

就是给一堆边，问能围成多少环，并查集检验一下就行了

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 1100 int pre[maxn], dis[maxn];int ok, ans;int find(int a){    return a == pre[a] ? a : pre[a] = find(pre[a]);}void merge(int x, int y){    int fx = find(x), fy = find(y);    if(fx != fy){        pre[fx] = fy;        return ;    }    ans++;}int main(){    int x, y, t, m, n;    while(~scanf("%d%d", &x, &y)){        for(int i = 0; i < x; i++){            pre[i] = i;        }        ans = 0;        for(int i = 0; i < y; i++){            scanf("%d%d", &m, &n);            merge(m, n);        }        printf("%d\n", ans);    }    return 0;}