POJ P2524 Ubiquitous Religions 基础并查集

Ubiquitous Religions

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 21066 Accepted: 10377

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 (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.

Input

The input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.

Output

For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.

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.

Source

Alberta Collegiate Programming Contest 2003.10.18

题意：学校里所有人都信宗教，每个人只能信一种宗教。

输入n个人m组数，每组数代表2个人信的是一种宗教。

问学校里存在多少种宗教。

简单并查集。初始存在n钟宗教。随着合并操作。宗教数会越来越少。以根节点人的标号代表宗教（教主即视感）。开一个score表示第1-n号宗教是否存在。最后扫一遍得到解。

POJ停电前留名。。。

#include <cstdio>int set[50010],score[50010];int m,n,x,y,i,no,ss;int FindSet(int x);void Union(int x,int y);int main(){no=0;while (scanf("%d%d",&n,&m) && n!=0){no++; ss=0;for (i=1;i<=n;++i){set[i]=i;score[i]=0;} for (i=1;i<=m;++i){scanf("%d%d",&x,&y);Union(x,y);}for (i=1;i<=n;++i){            FindSet(i);score[set[i]] = 1;        }for (i=1;i<=n;++i)if (score[i]) ss++;printf("Case %d: %d\n",no,ss);}return 0;}int FindSet(int x){if (x != set[x])set[x]=FindSet(set[x]);return set[x];}void Union(int x,int y){x=FindSet(x);y=FindSet(y);set[y] = x;return;}

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