TOJ3136

                                                      3136: Ubiquitous Religions

时间限制(普通/Java):2000MS/6000MS     内存限制:65536KByte
总提交: 274            测试通过:149

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.

样例输入

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

样例输出

Case 1: 1
Case 2: 7

提示

Huge input, scanf is recommended.

题解：

题目意思是两两结对，A认识B，B认识C，ABC就为同一群人，问最后有多少群人。

简单并查集。直接贴代码。

#include<stdio.h>
int fid[50010];
int find(int x)
{
if(x==fid[x])
{
return x;
}
else
{
return find(fid[x]);
}
}
void hebing(int a,int b)
{
a=find(a);
b=find(b);
if(a!=b)
{
fid[a]=b;
}
}
int main()
{
int n,m,i=0,j,a,b,s;
while(scanf("%d %d",&n,&m)!=EOF)
{
if(n==0&&m==0)break;
s=0;
i++;
for(j=0;j<n;j++)
{
fid[j]=j;
}
for(j=0;j<m;j++)
{
scanf("%d %d",&a,&b);
hebing(a,b);
}
for(j=0;j<=n;j++)
{
if(fid[j]==j)s++;
}
printf("Case %d: %d\n",i,s);
}
}