【杭电】[2120]Ice_cream's world I

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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
8 100 11 21 32 43 40 55 66 73 64 7
Sample Output
3
其实也就是找出成环个数已经在同一集合中的又进行合并find(x)==find(y)则形成一个环 res++
#include<stdio.h>int par[1200];int res;int find(int m) {    if(m==par[m])        return m;    else        return par[m]=find(par[m]);}void unite(int x,int y) {    x=find(x);    y=find(y);    if(x==y)        res++;    else {            par[y]=x;    }}int main() {    int n,m;    while(scanf("%d %d",&n,&m)!=EOF) {        for(int i=0; i<n; i++) {            par[i]=i;        }        res=0;        while(m--) {            int a,b;            scanf("%d %d",&a,&b);            unite(a,b);        }        printf("%d\n",res);    }    return 0;}
题目地址:【杭电】[2120]Ice_cream's world I


查看原文：http://www.boiltask.com/blog/?p=1978
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