HDU-1213 How Many Tables （并查集）

最新推荐文章于 2022-11-07 14:02:50 发布

原创最新推荐文章于 2022-11-07 14:02:50 发布 · 182 阅读

0 ·

CC 4.0 BY-SA版权

C++ 专栏收录该内容

28 篇文章

订阅专栏

本文介绍了一个经典的算法竞赛问题——最少餐桌数问题。在Ignatius的生日宴会上，他想知道至少需要多少张桌子才能让所有朋友坐在彼此认识的人旁边。通过使用并查集算法，文章详细解释了如何解决这个问题，包括输入解析、并查集初始化、查找和合并操作，最后输出所需最少的餐桌数量。

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52458 Accepted Submission(s): 26041

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

5 3

1 2

2 3

4 5

5 1

2 5

Sample Output

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

int pr[1005],n,m;
void init()
{
    for(int i=0;i<=n;i++)
        pr[i]=i;
}

int Find(int x)
{
    if(pr[x]!=x)
        pr[x]=Find(pr[x]);
    return pr[x];
}

void join(int x,int y)
{
    int a=Find(x),b=Find(y);
    if(a!=b)
        pr[b]=a;
    return ;
}


int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        init();
        int a,b,cnt=0;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            join(a,b);
        }
        for(int i=1;i<=n;i++)
        {
            if(Find(i)==i)
                cnt++;
        }
        printf("%d\n",cnt);
    }
    return 0;
}