HDU 1213 How Many Tables

最新推荐文章于 2025-09-10 22:21:01 发布

转载最新推荐文章于 2025-09-10 22:21:01 发布 · 48 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/ECJTUACM-873284962/p/6490747.html

文章标签：

#java #php

本文介绍了一个使用并查集算法解决的有趣问题：如何根据朋友间的熟识度最少分配多少张餐桌。通过实现并查集算法，解决了在特定条件下最小餐桌数目的计算问题。

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27651 Accepted Submission(s): 13726

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

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

分析：又是一波并查集，终于给我搞定了一道题，受了畅通工程的影响，这道题我也是刷刷刷的就过了！高兴！

具体并查集该怎么写，可以参照我的博客有关并查集专题讲解，绝对让你受益一生！

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int pre[1010];
 4 int t[1010];
 5 int find(int x)
 6 {
 7      int r=x;
 8      while(pre[r]!=r)
 9         r=pre[r];
10      int i=x,j;
11      while(pre[r]!=r)
12      {
13          j=pre[r];
14          pre[r]=r;
15          i=j;
16      }
17      return r;
18 }
19 void join(int x,int y)
20 {
21      int fx=find(x),fy=find(y);
22      if(fx!=fy)
23         pre[fy]=fx;
24 }
25 int main()
26 {
27      int N,M,a,b,i,ans,n,m;
28      while(scanf("%d",&N)!=EOF)
29      {
30          while(N--)
31          {
32              scanf("%d%d",&a,&b);
33              for(i=1;i<=a;i++)
34                 pre[i]=i;
35              for(i=1;i<=b;i++)
36              {
37                  scanf("%d%d",&n,&m);
38                  join(n,m);
39              }
40              memset(t,0,sizeof(t));
41              for(i=1;i<=a;i++)
42                 t[find(i)]=1;
43              for(ans=0,i=1;i<=a;i++)
44                 if(t[i])
45                 ans++;
46              printf("%d\n",ans);
47          }
48      }
49      return 0;
50 }