HDU 1213 How Many Tables 并查集路径压缩

最新推荐文章于 2021-08-08 13:18:43 发布

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最新推荐文章于 2021-08-08 13:18:43 发布

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分类专栏： ACM-图论文章标签：压缩并查集

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本文通过一个具体的生日派对场景问题，介绍了并查集的基本原理及其应用。文章详细解释了如何利用并查集来解决最少需要多少张桌子的问题，并提供了完整的代码实现。

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题目的链接如下：
http://acm.hdu.edu.cn/showproblem.php?pid=1213

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How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24459 Accepted Submission(s): 12241

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
本题的大致题意就是他要过生日，然后请了一些客人，让互相都认识的坐在一个桌子上，桌子可以无限大，然后问最多需要多少张桌子。
然后还有如果A认识B，B认识C那么A肯定认识C。
这题就是并查集的应用，把所有认识的人加到一个集合中，最后找有几个根节点即可，找根节点用if(father[i] == i）；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
using namespace std;
int father[1005];
void make_set()
{
    for(int i = 0 ; i < 1005 ; i++)
        father[i] = i;
}
/*
int find_root(int x)
{
    if(father[x] == x)
        return x;
    else return find_root(father[x]);
}
int Find(int x)
{

    if(father[x] != x)
        father[x] = Find(father[x]);
    return father[x];
}*/

int Find2(int x)
{
    int root = x;
    while(root != father[root])
        root = father[root];
    while(x != root){
        int tmp = father[x];
        father[root] = root;
        x = tmp;
    }
    return root;

}
int Union(int x, int y)
{
    int x_root, y_root;
    x_root = Find2(x);
    y_root = Find2(y);
    father[y_root] = x_root;
}

int main()
{
    int t;
    int a, b;
    int n, m;
    int cnt;
    cin >> t;
    while(t--){
        cin >> n >> m;
        make_set();
        cnt = 0;
        for(int i = 0 ; i < m ; i++){
            cin >> a >> b;
            Union(a,b);
        }
        //int now = father[1];
        //cnt++;
       for(int i = 1 ; i <= n ; i++)
            if(i == father[i])
                cnt++;
        cout << cnt << endl;
    }
    return 0;
}