How Many Tables||HDU1213

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本文介绍了一个经典的并查集应用案例，通过并查集算法来解决朋友间的分组问题，即最少需要多少张桌子才能让每个认识的人都坐在同一张桌子上。文章提供了一份完整的C++代码实现，并详细解释了如何通过并查集来确定不同朋友之间的关系。

题目来源于2017HPUACM暑期培训：https://cn.vjudge.net/contest/175491#problem/D

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

题解：就是并查集的运用，算出有几棵树

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int par[1005];
int tot;
int find(int x)//查找根节点 
{
    return x == par[x] ? x : /* */find(par[x]);
}
void unite(int x,int y)//合并 
{
    int fx,fy;
    fx=find(x);
    fy=find(y);
    if(fx!=fy)//判断两节点的根节点是否相同 
    {
        par[fy]=fx;//不相等就把x的根节点作为y根节点的父节点 
//      tot--;
    }
}
int main()
{
    int i,n,m,a,b,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
//      tot=n;
        tot=0;
        for(i=1;i<=n;i++)
           par[i]=i;
        while(m--)
        {
            scanf("%d%d",&a,&b);
            unite(a,b);
        }
////        if(tot<0) tot=0;
        for(i=1;i<=n;i++)
        {
            if(i==find(i))//计算有几个根节点，也就是几棵树 
              tot++;
        }
        printf("%d\n",tot);
    }
return 0;
}