HDU 1213 how many tables

最新推荐文章于 2022-11-07 14:02:50 发布

sky_loading

最新推荐文章于 2022-11-07 14:02:50 发布

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分类专栏：图论并查集

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本文介绍了一种使用并查集数据结构解决朋友分组问题的方法。在Ignatius的生日宴会上，需要根据相互认识的朋友进行最优分组，以确保每桌都是熟人。文章通过一个具体的例子说明了如何利用并查集来高效地计算最少所需的桌子数量。

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How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7850 Accepted Submission(s): 3813

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

Sample Output

2
4

Author

Ignatius.L

Source

杭电ACM省赛集训队选拔赛之热身赛

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Eddy

同样是一道裸的并查集，敲模板拿下

#include <stdlib.h> 
#include <iostream>
#include <algorithm>
using namespace std;
const int M = 1005;
int fam[M],member[M];
int cas,n,m,a,b;

void ufset( )
{
     for( int i=0 ; i<M ; i++ )
          member[i] = 1 , fam[i] = i;     
}

int find( int x )
{
    return x == fam[x] ? x : fam[x] = find( fam[x] );    
}

void merge( int a , int b )
{
     int x,y;
     x = find( a );
     y = find( b );
     if( x!=y ){
         fam[x] = y;
         member[y] += member[x];    
     }     
}

int main()
{
    scanf("%d",&cas);
    while( cas-- ){
           scanf("%d%d",&n,&m);
           ufset( ); 
           for( int i=0 ; i<m ; i++ ){
                scanf("%d%d",&a,&b);
                merge( a,b );
           }
           int sum = n;
           for( int i=1 ; i<=n ; i++ )
                if( fam[i] != i )
                    sum--;
           printf("%d\n",sum);
    }        
    return 0;
}