zoukankan      html  css  js  c++  java
  • Hdu 1213 How Many Tables

    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

    解题思路

      并查集;

    代码如下

     1 #include<iostream>
     2 using namespace std;
     3 int n, m, pre[10010];
     4 void init(){
     5     for(int i = 1; i <= n; i++)    pre[i] = i;
     6 }
     7 int getf(int v){
     8     if( pre[v] == v)    return v;
     9     else    return pre[v] = getf(pre[v]);
    10 }
    11 void meg(int v, int u){
    12     int t1 = getf(v), t2 = getf(u);
    13     if(t1 != t2)    pre[t2] = t1;
    14 }
    15 int main(){
    16     int t;
    17     cin >> t;
    18     while(t--){
    19         cin >> n >> m;
    20         init();
    21         int sum = 0;
    22         for(int i = 1; i <= m; i++){
    23             int x, y;
    24             cin >> x >> y;
    25             meg(x, y);
    26         }
    27         for(int i = 1; i <= n; i++){
    28             if(pre[i] == i){
    29                 sum++;
    30             }
    31         }
    32         cout << sum << endl;
    33     }
    34     return 0;
    35 }
    How Many Tables
  • 相关阅读:
    Windows照片查看器全屏浏览查看
    Windows调出软键盘
    IE叉事件
    对gridview的小改动
    gridview小把戏
    安全认证(转)
    用数组的方式实现DataTable中的distinct(转)
    TreeView的简单应用
    禁止按钮重复提交
    配置Microsoft Visual SourceSafe 2005的Internet访问(转)
  • 原文地址:https://www.cnblogs.com/zoom1109/p/11024898.html
Copyright © 2011-2022 走看看