HDU 1213 How Many Tables（并查集）

最新推荐文章于 2019-09-25 13:44:01 发布

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#each #output #integer #input #table #c

本文介绍了一种通过合并集合的方法解决派对中朋友座位安排问题的技术，确保每位朋友都能与认识的人同桌，提高派对体验。详细解释了算法逻辑及其实现过程。

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5194 Accepted Submission(s): 2452

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

第一次使用Find_Set的合并集合的方法，一次提交成功，内心受到小小的鼓舞。^_^

AC code：

#include <iostream>
using namespace std;

const int N=1001;
int flag[N];

int Find_Set(int x)
{
    if(x!=flag[x])
    {
        flag[x]=Find_Set(flag[x]);
    }
    return flag[x];
}

int main()
{
    int t,m,n,i,j,a,b,A,B,num;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        for(i=1;i<=n;i++) flag[i]=i;
        num=n;
        for(i=1;i<=m;i++)
        {
            cin>>A>>B;
            a=Find_Set(A);
            b=Find_Set(B);
            if(a>b)
            {
                flag[a]=b;
                num--;
            }
            else if(a<b)
            {
                flag[b]=a;
                num--;
            }
        }
        cout<<num<<endl;
    }
    return 0;
}