HDU1213 How Many Tables (并查集)

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探讨如何在考虑人际关系的前提下，计算最少需要多少张桌子来安排聚会。通过输入的朋友数量和已知相互认识的朋友对，使用并查集算法确定最少桌数。

Today is Ignatius’ birthday. He invites a lot of friends. Now it’s dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.
Input
The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.
Output
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
Sample Input

Sample Output

2
4

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
#define MAXN 50050
int pre[30030];
int find(int x)
{
    int r = x;      //用r来传递信息
    while(pre[r] != r)  //如果r的上级不是自身
        r = pre[r];     //r就接着找上级，直到找到老大
    return r;           //告诉你x的老大是谁
}

void join(int x, int y)     //x，y成为朋友，于是他们两个的联盟合并了
{
    int fx = find(x);       //找x的老大
    int fy = find(y);       //找y的老大
    if(fx != fy){
        pre[fx] = fy;       //x的老大成了y老大的小弟
    }
}

int main()
{
    int n, m, i, j, a, b;
    cin >> n;
    while(n--)
    {
        int sum = 0;
        cin >> a >> b;
        for(j=1; j<=a; j++){
            pre[j] = j;
        }
        while(b--){
            int x, y;
            cin >> x >> y;
            join(x, y);


        }
        for(j=1; j<=a; j++){
            if(pre[j] == j)
                sum++;
        }
        cout << sum << endl;
        if(n>1)
            getchar();
    }
    return 0;
}