【杭电】[2120]Ice_cream's world I
2016-08-01 16:31
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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 10 0 1 1 2 1 3 2 4 3 4 0 5 5 6 6 7 3 6 4 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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