How Many Tables
题目描述：
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
思路：
简单的寻找根节点，并合并根节点，集合成一个集合就行了，简单的并查集模板题。
代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;
int f[2000];
int p[2000];
int LemonFind(int u)//找出根节点
{
if(f[u]!=u)
{
f[u]=LemonFind(f[u]);
}
return f[u];
}
void LemonGetroot(int u,int v)//合并集合
{
int x1=LemonFind(u);//找出根节点
int x2=LemonFind(v);//找出根节点
if(x1!=x2)//如果根节点不相等,就改变父节点的去向
{
f[x2]=x1;
}
}
int main()
{
int t;
int n,m;
int u,v;
scanf("%d",&t);
while(t--)
{
set<int>q;
int cnt=0;
scanf("%d %d",&n,&m);
for(int i=1;i<=n;i++)//初始化一下根节点
{
f[i]=i;
}
for(int i=1;i<=m;i++)
{
scanf("%d %d",&u,&v);
LemonGetroot(u,v);//合并集点
}
for(int i=1;i<=n;i++)
{
q.insert(LemonFind(i));//用set来判断重合元素
}
printf("%d\n",q.size());
}
}