POJ 2524 Ubiquitous Religions (并查集)
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题目
Description
There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in.
You know that there are n students in your university
Input
The input consists of a number of cases. Each case starts with a line specifying the integers
Output
For each test case, print on a single line the case number (starting with
Sample Input
10 91 21 31 41 51 61 71 81 91 1010 42 34 54 85 80 0
Sample Output
Case 1: 1Case 2: 7
Hint
Huge input, scanf is recommended.
分析
并查集模板题,不多说,直接上代码
代码
#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<queue>#include<stack>#include<vector>#include<cmath>#include<set>#include<map>#include<cstdlib>#include<functional>#include<climits>#include<cctype>#include<iomanip>using namespace std;typedef long long ll;#define INF 0x3f3f3f3f#define mod 1e9+7#define clr(a,x) memset(a,x,sizeof(a))const double eps = 1e-6;int religion[1000000];int find(int a){ if(religion[a]!=a) religion[a]=find(religion[a]); return religion[a];}void merge(int a,int b){ int fx,fy; fx=find(a); fy=find(b); if(fx!=fy) religion[fx]=fy;}int main(){ int a,b; int t,n; int kase=0; while(scanf("%d%d",&t,&n)==2,t+n) { int cnt=0; clr(religion,0); for(int i=1;i<=t;i++) { religion[i]=i; } for(int i=0;i<n;i++) { scanf("%d%d",&a,&b); merge(a,b); } for(int i=1;i<=t;i++) { if(find(i)==i) cnt++; } printf("Case %d: %d\n",++kase,cnt); } return 0;}
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