Ubiquitous Religions
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 39239 | Accepted: 18708 |
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 分析：并查集模板题。
#include#include #include using namespace std;int M,N,cas=0;//N个点M条边 int pre[50010];//pre[i]=x表示点i的父亲节点为点x void init()//初始化N个元素pre[x]=x,即开始时有N个集合 { for(int i=0;i<=N;i++) pre[i]=i;}int find(int x)//查询某点所在树的根 { int root=x; while(pre[root]!=root) root=pre[root]; int i=x,k; while(k!=root)//路径压缩,让x到根经过的所有节点直接指向根节点 { k=pre[i]; pre[i]=root; i=k; } return root;}int unite(int x,int y){ int rootx=find(x),rooty=find(y); if(rootx!=rooty)//合并 {pre[rootx]=pre[rooty];return 1;} return 0;}int main(){ while(scanf("%d%d",&N,&M)) { init(); if(!M&&!N) break; int x,y,ans=N; while(M--) { scanf("%d%d",&x,&y); ans-=unite(x,y); } printf("Case %d: %d\n",++cas,ans); } return 0;}