POJ 2524 Ubiquitous Religions 并查集

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解题思路

题意：
问一个大学里学生的宗教，通过问一个学生可以知道另一个学生是不是跟他信仰同样的宗教。问学校里最多可能有多少个宗教。
也就是给定一个图的点数和相应的边，问有多少个连通分量。

思路：
使用并查集，输入时使每个节点的根节点都指向自己，枚举每条边，把根节点不相同的合并。完后列举每个节点，把每个节点的根节点放在数组arr里，arr用qsort排序，数出有多少个不相同的根节点，就是结果。本来合并的时候没有用到rank，700多MS，用之后减少到500多。所以并查集里面那些优化都一定要用啊。

题目描述

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.

输入

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.

输出

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.

示例输入

示例输出

Case 1: 1
Case 2: 7#include<stdio.h>
int s[100002];
int find(int x)
{int r=x;while(r!=s[r])r=s[r];return r;
}void  merg(int a,int b)
{int fa,fb;fa=find(a);fb=find(b);if(fa!=fb)s[fa]=fb;
}int main()
{int m,n,a,b,i,flag,Case=1;while(~scanf("%d %d",&n,&m),m||n){flag=0;for(i=1; i<=n; i++)s[i]=i;while(m--){scanf("%d %d",&a,&b);merg(a,b);}for(i=1;i<=n;i++){if(s[i]==i)flag++;}printf("Case %d: %d\n",Case++,flag);}return 0;
}

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