How Many Tables(并差集)

今天药忘吃喽~ 2022-02-25 06:14 180阅读 0赞

How Many Tables

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

核心代码为：

for(int i=1;i<=m;i++)
    	{
    		if(mark[pre[find(i)]]++==0)//判断是否在一个集合中 不在一个集合中 那么ans++; 
    		ans++;
    	}

这个是什么意思来？  
事先给mark\[\]数组赋初始值为0,然后每一个人都要遍历判断一下自己所在的上级哪个没有出现，那么那个人就得自己一张桌子了。例如：  
5 3  
1 2  
2 3  
3 4  
pre\[1\]=1; pre\[2\]=1; pre\[3\]=2; pre\[4\]=3;  
mark\[1\]=2; mark\[2\]=1; mark\[3\]=1; **mark\[4\]=0;mark\[5\]=0**  
所以要准备两张桌子

#include <cstdio>
    #include <iostream>
    #include <cstring>
    #define N 10000
    int pre[N],mark[N];
    using namespace std;
    
    int find(int x)
    {
    	int r=x;
    	while(pre[r]!=r)
    	{
    		r=pre[r];
    	}
    	
    	int i=x,j;
    	while(i!=r)
    	{
    		j=pre[i];
    		pre[i]=r;
    		i=j;
    	}
    	return r;
    }
    
    void init()
    {
    	for(int i=1;i<=N;i++)
    	{
    	pre[i]=i;
    	}
    	memset(mark,0,sizeof(mark));
    }
    
    int main()
    {
    	int t,m,n,p1,p2;
    	scanf("%d",&t);
    	while(t--)
    	{
    	init();
    	int ans=0;
    	scanf("%d %d",&m,&n);
    	while(n--)
    	{
    		scanf("%d %d",&p1,&p2);
    		int fx=find(p1);
    		int fy=find(p2);
    		if(fx!=fy)
    		pre[fx]=fy;	
    	}	
    	for(int i=1;i<=m;i++)
    	{
    		if(mark[pre[find(i)]]++==0)//判断是否在一个集合中 不在一个集合中 那么ans++; 
    		ans++;
    	}
    	printf("%d\n",ans);
    	}
    	return 0;
    }