HDU1213--How Many Tables

最新推荐文章于 2025-08-23 19:36:46 发布

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原文链接：http://www.cnblogs.com/zafuacm/p/3199697.html

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#java

本文介绍了一个关于朋友分组的经典并查集问题，通过解决Ignatius生日派对上朋友如何就座的问题来讲解并查集算法的应用。该问题要求最小化餐桌数量，同时确保熟识的朋友能坐在同一桌。文章提供了完整的C++代码实现，并详细解释了并查集的初始化、查找及合并操作。

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How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9097 Accepted Submission(s): 4448

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

Author
Ignatius.L

Source
杭电ACM省赛集训队选拔赛之热身赛

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简单并查集

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<cmath>
 6 using namespace std;
 7 
 8 int parent[1001];
 9 int n,m;
10 
11 void Ufset()
12 {
13     int i;
14     for(i=1;i<=n;i++)
15     {
16         parent[i]=-1;
17     }
18 }
19 
20 int Find(int x)
21 {
22     int s;
23     for(s=x;parent[s]>0;s=parent[s]){}
24     while(s!=x)
25     {
26         int temp=parent[x];
27         parent[x]=s;
28         x=temp;
29     }
30     return s;
31 }
32 
33 void Union(int R1,int R2)
34 {
35     int r1=Find(R1),r2=Find(R2);
36     int temp=parent[r1]+parent[r2];
37     if(parent[r1]>parent[r2])
38     {
39         parent[r1]=r2;
40         parent[r2]=temp;
41     }
42     else
43     {
44         parent[r2]=r1;
45         parent[r1]=temp;
46     }
47 }
48 
49 int main()
50 {
51     int t;
52     scanf("%d",&t);
53     while(t--)
54     {
55         scanf("%d%d",&n,&m);
56         Ufset();
57         while(m--)
58         {
59             int u,v;
60             scanf("%d%d",&u,&v);
61             if(Find(u)!=Find(v))
62             {
63                 Union(u,v);
64             }
65         }
66         int sum=0;
67         int ans=0;
68         for(int i=1;i<=n;i++)
69         {
70             if(abs(parent[i])>1&&parent[i]<0)
71             {
72                 sum++;
73                 ans+=abs(parent[i]);
74             }
75         }
76         if(ans==n)
77             printf("%d\n",sum);
78         else
79             printf("%d\n",sum+n-ans);
80     }
81     return 0;
82 }