HDU 1213 How Many Tables（并查集）

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探讨了如何通过并查集算法解决一个生日派对中朋友间的座位安排问题，确保熟识的朋友能坐在同一桌，而陌生人则不会被安排在一起。

How Many Tables

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Problem Description

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

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int mp[1111];
void init()
{
    for(int i=0;i<1111;i++)
        mp[i]=i;
}
int find(int x)
{
    if(mp[x]!=x)
        mp[x]=find(mp[x]);
    return mp[x];
}
void mix(int a,int b)
{
    int aa=find(a);
    int bb=find(b);
    if(aa!=bb)
        mp[aa]=mp[bb];
}
int main()
{
    int T;
    int n,m;
    int u,v;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        init();
        for(int i=0;i<m;i++)
        {
            cin>>u>>v;
            mix(u,v);
        }
        int res=0;
        for(int i=1;i<=n;i++)
        {
            if(mp[i]==i)
                res++;
        }
        cout<<res<<endl;
    }
    return 0;
}