HDU - 1213 - How Many Tables（并查集）

#HDU - 1213 - How Many Tables
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
2
5 3
1 2
2 3
4 5

5 1
2 5
Sample Output
2
4
题目链接
这个题目就是告诉你一个人数，然后再告诉你里面人的关系，a 跟 b 认识， b 跟 c 认识， 那么a 跟 c也认识，只有互相认识的人可以做一个桌子，谁都不认识就自己一张桌子。问你到最后应该用多少张桌子。

其实就是个裸的并查集，把所有互相认识的人都并到一起，最后查找有多少个是父节点就可以了。
AC代码：

#include<stdio.h>

int f[1010];

void init()
{
    for(int i = 1; i <= 1010; i++)
        f[i] = i;
}

int getf(int x)
{
    if(f[x] == x)
        return x;
    else
        return f[x] = getf(f[x]);
}

void unit(int a, int b)
{
    int x = getf(a), y = getf(b);
    f[y] = x;
}

int main()
{
    int T, n, m;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        init();

        for(int i = 0; i < m; i++)
        {
            int a, b;
            scanf("%d%d", &a, &b);
            unit(a, b);
        }
        int num = 0;
        for(int i = 1; i <= n; i++)
            if(i == f[i])
                num++;
        printf("%d\n", num);
    }
    return 0;
}