zoukankan      html  css  js  c++  java
  • hdu_1213_How Many Tables_201403091126

    How Many Tables

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 11038    Accepted Submission(s): 5447


    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
     
     1 #include <stdio.h>
     2 int pre[1010];
     3 int find(int x)
     4 {
     5     int i,r,t;
     6     t=r=x;
     7     while(r!=pre[r])
     8     {
     9         r=pre[r];
    10     }
    11     while(t!=r)
    12     {
    13         i=pre[t];
    14         pre[t]=r;
    15         t=i;
    16     }
    17     return r;
    18 }
    19 int main()
    20 {
    21     int T;
    22     scanf("%d",&T);
    23     while(T--)
    24     {
    25         int m,n,i,j,a,b,pa,pb,total;
    26         scanf("%d %d",&n,&m);
    27         total=n;
    28         for(i=1;i<=n;i++)
    29         pre[i]=i;
    30         for(i=0;i<m;i++)
    31         {
    32             scanf("%d %d",&a,&b);
    33             pa=find(a);
    34             pb=find(b);
    35             if(pa!=pb)
    36             {
    37                 pre[pb]=pa;
    38                 total--;
    39             }
    40         }
    41         printf("%d
    ",total);
    42     }
    43     return 0;
    44 }

    //并查集简单应用

  • 相关阅读:
    透视表提取不反复记录(1)-出现值
    ORA-38760: This database instance failed to turn on flashback database
    Android蓝牙串口程序开发
    指尖上的电商---(5)schema.xml配置具体解释
    iOS-UIImage imageWithContentsOfFile 和 imageName 对照
    JSON-RPC轻量级远程调用协议介绍及使用
    POJ 2296 Map Labeler(2-sat)
    接口測试-HAR
    [Leetcode]Combination Sum II
    MarkDown、Vim双剑合璧
  • 原文地址:https://www.cnblogs.com/xl1027515989/p/3589734.html
Copyright © 2011-2022 走看看