How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22498 Accepted Submission(s): 11193
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
2
5 3
1 2
2 3
4 5
5 1
2 5
Sample Output
2
4
题解：简单的并查集（集合的合并），有关系的人，即属于同一个集合的人一桌！看有多少棵树？
代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
const int maxx=1005;
int par[maxx];
int n,m;
void init()
{
for(int i=1;i<=n;i++)
par[i]=i;
}
int find(int x)
{
if(par[x]==x)
return x;
else
return par[x]=find(par[x]);
}
void unite(int x,int y)
{
x=find(x);
y=find(y);
if(x==y)
return ;
par[x]=y;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init();
for(int i=1;i<=m;i++)
{
int x,y;
scanf("%d %d",&x,&y);
unite(x,y);
}
int sum=0;
for(int i=1;i<=n;i++)
{
if(par[i]==i)
sum++;
}
printf("%d\n",sum);
}
return 0;
}