HDU 1213-How Many Tables

最新推荐文章于 2019-03-18 20:48:45 发布

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最新推荐文章于 2019-03-18 20:48:45 发布

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分类专栏： ACM-并查集

本文链接：https://blog.youkuaiyun.com/lz161530245/article/details/77773035

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本文介绍了一个经典的并查集应用案例，通过Ignatius的生日聚会问题，探讨了如何利用并查集来确定最少的分组数量，确保每个组内成员互相认识。文章详细解释了并查集的实现过程，并提供了完整的代码示例。

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How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33870 Accepted Submission(s): 16941

题目连接：点击打开链接

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

题意：

Ignatius过生日时邀请了很多朋友，到晚饭时间，他想知道至少需要多少桌子，他的朋友坐桌子有个条件，不和陌生人坐一起，因为Ignatius的朋友并不是彼此都认识，但是呢，朋友的朋友也算是自己的朋友，什么意思呢，比如说，A和B是朋友，B和C是朋友，E和F是朋友，所以就需要至少两个桌子，即A，B，C三人坐一个桌子，E，F坐一个桌子。

分析：

考察并查集的基本应用。

#include<iostream>
#include<stdio.h>
#include<cmath>
using namespace std;
int pre[1010],s[1010];
int Find(int b)
{
    if(pre[b]==b)
        return b;
    else
        return Find(pre[b]);
}
void pei(int jia,int yi)
{
    int f1=Find(jia);
    int f2=Find(yi);
    if(f1!=f2)
    {
        pre[f1]=f2;
    }
}
int main()
{
    int n,m,a,jia,yi,sum;
    scanf("%d",&a);
    while(a--)
    {
        scanf("%d %d",&n,&m);
        for(int i=1;i<=n;i++)
            pre[i]=i;
         while(m--)
         {
            scanf("%d %d",&jia,&yi);
            pei(jia,yi);
        }
        sum=0;
        for(int i=1;i<=n;i++)
        {
             if(Find(i)==i)
                sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}