How Many Tables HDU - 1213 （并查集）

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

题意理解：此题其实就是并查集的裸题，最开始我还在想会不会每个桌子会有人数限制，但读完题后发现，这个桌子也是无限大的，所以只要判断它有几个连通块就可以了。

代码：

#include 
#include 
#include 
#include 
using namespace std;

int pre[1005];

int Find(int root)
{
    int son, temp;
    son = root;
    while(root != pre[root])
        root = pre[root];
    while(son != root)  // 路径压缩
    {
        temp = pre[son];
        pre[son] = root;
        son = temp;
    }
    return root;
}

void Join(int root1, int root2)
{
    int x = Find(root1);
    int y = Find(root2);
    if(x != y)
        pre[x] = y;
}

int main()
{
    int T, N, M;
    cin >> T;
    while(T--)
    {
        int a, b;
        set Sum;
        cin >> N >> M;
        for(int i = 1; i < 1005; i++)
            pre[i] = i;
        while(M--)
        {
            cin >> a >> b;
            Join(a, b);
        }
        for(int i = 1; i <= N; i++)
        {
            int t = Find(i);
            Sum.insert(t);
        }
        printf("%d\n",Sum.size() );
    }
    return 0;
}