zoukankan      html  css  js  c++  java
  • hdu 1213 How Many Tables (并查集)

    How Many Tables
    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 44432    Accepted Submission(s): 22199

    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

    C/C++:

     1 #include <map>
     2 #include <queue>
     3 #include <cmath>
     4 #include <string>
     5 #include <cstdio>
     6 #include <cstring>
     7 #include <climits>
     8 #include <algorithm>
     9 #define INF 0xffffff
    10 using namespace std;
    11 
    12 int n, m, my_pre[1010];
    13 
    14 int my_find(int x)
    15 {
    16     int temp = x;
    17     while (temp != my_pre[temp]) temp = my_pre[temp];
    18     int i = x, j;
    19     while (temp != my_pre[i])
    20     {
    21         j = my_pre[i];
    22         my_pre[i] = temp;
    23         i = j;
    24     }
    25     return temp;
    26 }
    27 
    28 int main()
    29 {
    30     int t;
    31     scanf("%d", &t);
    32 
    33     while (t --)
    34     {
    35         int my_cnt = 1, my_first_temp;
    36         scanf("%d%d", &n, &m);
    37         for (int i = 1; i <= n; ++ i)
    38             my_pre[i] = i;
    39 
    40         for (int i = 1; i <= m; ++ i)
    41         {
    42             int a, b, n1, n2;
    43             scanf("%d%d", &a, &b);
    44             n1 = my_find(a), n2 = my_find(b);
    45             if (n1 != n2)
    46                 my_pre[n1] = n2;
    47         }
    48 
    49         my_first_temp = my_find(1);
    50         for (int i = 2; i <= n; ++ i)
    51             if (my_find(i) != my_first_temp)
    52                 my_pre[my_find(i)] = my_first_temp, ++ my_cnt;
    53 
    54         printf("%d
    ", my_cnt);
    55     }
    56     return 0;
    57 }
  • 相关阅读:
    一个切换鼠标左右手的小工具
    聊一聊高并发高可用那些事
    PHP 笔记
    PHP 安装 XDebug
    php symfony/var-dumper 打印插件
    PHP字符串和数组
    ThinkPHP自定义分页模板
    匿名函数
    SQL优化工具
    spring cloud 服务容错保护
  • 原文地址:https://www.cnblogs.com/GetcharZp/p/9445711.html
Copyright © 2011-2022 走看看