hdu 1213 How Many Tables

并查集解决社交派对桌次安排问题

最新推荐文章于 2025-09-12 18:08:49 发布

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最新推荐文章于 2025-09-12 18:08:49 发布

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原文链接：http://www.cnblogs.com/jeff-wgc/p/4449360.html

本文介绍如何使用并查集算法解决派对中嘉宾座位安排问题，确保每位嘉宾都能与认识的人坐在一起，从而提高派对体验。通过简单的数据输入和输出示例，展示了算法的应用过程。

http://acm.hdu.edu.cn/showproblem.php?pid=1213

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16666 Accepted Submission(s): 8167

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

分析：

简单的并查集，把有关系的嘉宾放到一个节点下（集合），选择一个嘉宾作为父亲节点，最后只判断有多少个不同的父亲节点即可。

AC代码：

 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <string.h>
 5 #include <string>
 6 #include <math.h>
 7 #include <stdlib.h>
 8 #include <queue>
 9 #include <stack>
10 #include <set>
11 #include <map>
12 #include <list>
13 #include <iomanip>
14 #include <vector>
15 #pragma comment(linker, "/STACK:1024000000,1024000000")
16 #pragma warning(disable:4786)
17 
18 using namespace std;
19 
20 const int INF = 0x3f3f3f3f;
21 const int MAX = 1000 + 10;
22 const double eps = 1e-8;
23 const double PI = acos(-1.0);
24 
25 int a[MAX];
26 
27 int find(int x)
28 {
29     while(x != a[x])
30         x = a[x];
31     return x;
32 }
33 
34 int main()
35 {
36     int T;
37     scanf("%d",&T);
38     {
39         while(T --)
40         {
41             int n , m;
42             scanf("%d %d", &n , &m);
43             int i;
44             for(i = 0;i <= n;i ++)
45                 a[i] = i;
46             int temp1 , temp2;
47             while(m --)
48             {
49                 scanf("%d %d",&temp1 , &temp2);
50                 int fa = find(temp1);
51                 int fb = find(temp2);
52                 if(fa != fb)
53                     a[fa] = fb;
54             }
55             int ans = 0;
56             for(i = 1;i <= n;i ++)
57                 if(a[i] == i)
58                     ans ++;
59             printf("%d\n",ans);
60         }
61     }
62     return 0;
63 }