HDU 1213-How Many Tables（简单并查集）

How Many Tables

[b]Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38275    Accepted Submission(s): 19100
[/b]
原题网址：http://acm.hdu.edu.cn/showproblem.php?pid=1213
[align=left]Problem Description[/align]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.
 
[align=left]Input[/align]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.
 
[align=left]Output[/align]For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
 
[align=left]Sample Input[/align]

25 31 22 34 55 12 5 [align=left]Sample Output[/align]

24 
[align=left]Author[/align]Ignatius.L 
[align=left]Source[/align] 杭电ACM省赛集训队选拔赛之热身赛

题目大意：一个人生日，要请他朋友吃饭，他有很多朋友，他要为他们认识的朋友准备桌子吃饭，相互认识的可以坐在一桌，不认识的分开坐，求出最小所需要的桌子数目。解题思路：这里我们需要写一个并查集来储存他们所认识的朋友，相互认识放在一棵树上，算出树的数量即可！
AC代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<time.h>
#include<map>
#include<string>
#include<algorithm>
#include<set>
#define N 1010

using namespace std;

int table
,sum;

int formalset(int m) //初始化他们每个朋友为一棵独立的树，且树根为他们自己
{
for(int i = 1; i <= m ;i ++)
{
table[i] = i;
}
}

int findtable(int k) //查找树根（初学者一定要画图模拟！）
{
if(table[k] != k)
table[k] = findtable(table[k]);
return table[k];
}

void isfriend(int a, int b) //这两个朋友认识，把他们合并同一个树根（这个请自己画图模拟一下，不然很难理解！）
{
if(findtable(a) == findtable(b)) //是朋友并且已经合并到一起过了，直接返回
return;
sum --; //是朋友，没有合并，桌子数减一（开始桌子数为朋友总数）
table[findtable(b)] = findtable(a); //b和a的根合并为a的根
}

int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int m, n, a, b;
scanf("%d%d", &m, &n);
sum = m;
formalset(m);
for(int i = 0; i < n; i ++)
{
scanf("%d%d", &a, &b);
isfriend(a, b);
}
printf("%d\n",sum);
}
return 0;
}