poj2524——Ubiquitous Religions【并查集基础】

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本文介绍了一道关于并查集的基础题目，通过询问学生之间的宗教信仰相似性来估算校园内不同宗教信仰的数量上限。利用并查集算法进行数据处理，实现了高效地合并同类项，并最终计算出可能的宗教类别数量。

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Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 38776		Accepted: 18486

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

Sample Output

Case 1: 1
Case 2: 7

Hint

Huge input, scanf is recommended.

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题目大意：给你一个数N，m,输入m组数据，每组数据有两个数，把两组数据归为一类，问最后1~n中有多少组数据属于同一类（单独一组数据自成一类）。

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
const ll MAX_N=100000;//数组不要阔的太大，不然会内存超限，之前就因为数组阔的太大wa了
ll par[MAX_N],rank[MAX_N],a[MAX_N],b[MAX_N],c[MAX_N];
void init(ll n){
	for(ll i=0;i<n;i++){
		par[i]=i;
		rank[i]=0;
	}
}
ll find(ll x){
	if(par[x]==x){
		return x;
	}else{
		return par[x]=find(par[x]);
	}
}
void unite(ll x,ll y){
	x=find(x);
	y=find(y);
	if(x==y) return;
	if(rank[x]<rank[y]){
		par[x]=y;
	}else{
		par[x]=y;
		if(rank[x]==rank[y]) rank[x]++;
	}
}
bool same(ll x,ll y){
	return find(x)==find(y);
}
ll count_sets(ll n){
	ll cnt=0;
	for(ll i=1;i<=n;i++){
		if(find(i)==i) cnt++;
	}
	return cnt;
}//统计并查集中集合的个数 
int main(){
	ll n,m,ans,kase=1;
	while(scanf("%lld %lld",&n,&m)&&n,m){
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(c,0,sizeof(c));
		ll i,j;
		init(n);//初始化并查集 
		//for(i=1;i<=n;i++) c[i]=i; 
		for(i=0;i<m;i++){
			scanf("%lld %lld",&a[i],&b[i]);
			/*c[a[i]]=0;
			c[b[i]]=0;*/
			unite(a[i],b[i]);//合并为统一集合 
		}
		ans=count_sets(n)+1;
		printf("Case %lld: %lld\n",kase++,ans);
	}
	return 0;
}