poj2524 解题报告

基于并查集的一道简单题目

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 22334		Accepted: 10996

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

本题要计算一个小组中最多有多少种不同的信仰，只需用小组的人数减去已有小组内人数的总数，再加上小组的数目，就得到了结果。
如果两个节点不在一个小组，则要合并他们，用MaxNum-2+1，MaxNum减去小组内两个人，还要加1的小组数目，所以只需要在合并两个不同节点时用MaxNum--即可。

代码：

 #include <iostream>
 using namespace std;

 int father[], num[];
 int count = ;
 int MaxNum;

 void MakeSet(int x)
 {
     father[x] = x;
     num[x] = ;
 }

 int FindSet(int x)
 {
     if(x != father[x])
         father[x] = FindSet(father[x]);
     return father[x];
 }

 void Union(int x, int y)
 {
     int GrandX = FindSet(x);
     int GrandY = FindSet(y);

     if(GrandX == GrandY)
         return;
     if(num[GrandX] < num[GrandY])
     {
         father[GrandX] = GrandY;
         num[GrandY] += num[GrandX];
     }
     else
     {
         father[GrandY] = GrandX;
         num[GrandX] += num[GrandY];
     }
     MaxNum--;
 }

 int main()
 {
     int n, m;
     int Case = ;

     while((cin >> n, cin >> m) && !(m== && n==))
     {
         MaxNum = n;
         for(int i = ; i < n; i++)
             MakeSet(i);
         for(int i = ; i < m; i++)
         {
             int x, y;
             cin >> x >> y;
             Union(x, y);
         }
         cout << "Case " << ++Case << ": " << MaxNum << endl;
     }
 }