标签：poj 幷查集
Ubiquitous Religions
| Time Limit: 5000MS |
|
Memory Limit: 65536K |
| Total Submissions: 23090 |
|
Accepted: 11378 |
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 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
Huge input, scanf is recommended.
Source
#include <iostream>
#include <cstdio>
using namespace std;
int f[50005],sum;
//查找x的父节点，如果f[x]==x，他的父节点就是本身。
int find (int x)
{
if(f[x]!=x)
f[x]=find(f[x]);
return f[x];
}
//如果不是一个集合（父节点不相同），合并集合（将父节点改成b的父节点）。
void make(int a,int b)
{
int f1=find(a);
int f2=find(b);
if(f1!=f2)
{
f[f2]=f1;
sum--; //宗教数减一。
}
}
int main()
{
int n,m,p=1,i;
while(scanf("%d%d",&n,&m)!=EOF)
{
if(n==0&&m==0) break;
for(i=1;i<=n;i++)
f[i]=i;
sum=n;
for(i=1;i<=m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
make(a,b);
}
printf("Case %d: %d\n",p++,sum);
}
return 0;
}
POJ 2524 Ubiquitous Religions (幷查集),布布扣,bubuko.com
POJ 2524 Ubiquitous Religions (幷查集)
标签：poj 幷查集
原文地址：http://blog.csdn.net/qq2256420822/article/details/37763809
