hdu 1213 How Many Tables（并查集的简单应用）

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16333    Accepted Submission(s): 8012

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

 

最近在专攻并查集的题目，读完第一段就基本确认又是一个并查集。。。分析过程就不给了，英文还过关的话读一遍就能懂。

AC代码：

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std ;

int f[1005] ;
int t,m,n ;
int x,y ;

void init()
{
    for(int i = 1 ;i<=n ;i++)
    {
        f[i] = i ;
    }
}

int getRoot(int v)
{
    if(f[v] == v)
       return v ;
    else
    {
        f[v] = getRoot(f[v]) ;
        return f[v] ;
    }
}

void merge(int v,int u)
{
    int t1 = getRoot(v) ;
    int t2 = getRoot(u) ;
    if(t1 != t2)
    {
        f[t2] = t1 ;
    }
}

int main()
{
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            scanf("%d%d",&n,&m) ;
            init() ;
            for(int i = 0 ;i<m ;i++)
            {
                scanf("%d%d",&x,&y) ;
                merge(x,y) ;
            }
            int sum = 0 ;
            for(int i = 1 ;i<=n ;i++)
            {
                if(f[i] ==i)
                    sum++ ;
            }
            printf("%d
",sum) ;
        }
    }
    return 0 ;
}