sdnu-并查集-weeklyexam ——G - How Many Tables

最新推荐文章于 2020-09-01 20:25:53 发布

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本文介绍了一个简单的问题：如何确定最少需要多少张桌子来安排一场生日派对，确保每位宾客都能与熟识的人同桌。通过使用并查集算法解决这一问题，实现了高效地计算所需的桌子数量。

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

来一个比较简单的题歇一会~~

输入人员关系看看有几个部落

常规的并查集问题

#include <cstdlib>   
#include <iostream>  
#include <algorithm>  
using namespace std;  
const int maxa = 1005;  
int pre[maxa];  
int t,n,m,a,b;  
int find( int x )  
{  
    return x == pre[x] ? x : pre[x] = find( pre[x] );      
}  
  
void join( int a , int b )  
{  
     a = find( a );  
     b = find( b );  
     if( a != b )
	 {  
         pre[a] = b;     
     }       
}  
  
int main()  
{  
    scanf("%d",&t);  
    while( t-- )
	{  
           cin>>n>>m;   
		 for( int i=0 ; i<maxa ; i++ )  
		 {
		  	pre[i] = i;  
		 }
           for( int i=0 ; i<m ; i++ )
		   {  
                cin>>a>>b; 
               join(a,b);  
           }  
           int sum = n;  
           for( int i=1 ; i<=n ; i++ ) 
		   {
			   	if( pre[i] != i )  
	                {
	                	sum--; 
					}            
			} 
			cout<<sum<<endl;
                
    }          
    return 0;  
}