Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 23997 | Accepted: 11807 |
Description
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
Output
Sample Input
10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0
Sample Output
Case 1: 1
Case 2: 7
Hint
Source
#include <iostream>
#include <stdio.h>
using namespace std; #define MAXN 50010 int par[MAXN]; //par[x]表示x的父节点
int n; void Init() //初始化
{
int i;
for(i=;i<=n;i++)
par[i] = i;
} int Find(int x) //查询x的根节点并路径压缩
{
if(par[x]!=x)
par[x] = Find(par[x]);
return par[x];
} void Union(int x,int y) //合并x和y所在集合
{
par[Find(x)] = Find(y);
} int main()
{
int m,x,y,i,Case=;
while(scanf("%d%d",&n,&m)!=EOF){
if(n== && m==)
break;
//初始化
Init();
//m次询问
while(m--){
scanf("%d%d",&x,&y);
Union(x,y);
}
int ans = n;
for(i=;i<=n;i++) //统计
if(par[i]!=i)
--ans;
printf("Case %d: %d\n",Case++,ans);
}
return ;
}
Freecode : www.cnblogs.com/yym2013