zoukankan      html  css  js  c++  java
  • HDU 4883 TIANKENG’s restaurant

                TIANKENG’s restaurant



     

    Problem Description
    TIANKENG manages a restaurant after graduating from ZCMU, and tens of thousands of customers come to have meal because of its delicious dishes. Today n groups of customers come to enjoy their meal, and there are Xi persons in the ith group in sum. Assuming that each customer can own only one chair. Now we know the arriving time STi and departure time EDi of each group. Could you help TIANKENG calculate the minimum chairs he needs to prepare so that every customer can take a seat when arriving the restaurant?
     

     

    Input
    The first line contains a positive integer T(T<=100), standing for T test cases in all.

    Each cases has a positive integer n(1<=n<=10000), which means n groups of customer. Then following n lines, each line there is a positive integer Xi(1<=Xi<=100), referring to the sum of the number of the ith group people, and the arriving time STi and departure time Edi(the time format is hh:mm, 0<=hh<24, 0<=mm<60), Given that the arriving time must be earlier than the departure time.

    Pay attention that when a group of people arrive at the restaurant as soon as a group of people leaves from the restaurant, then the arriving group can be arranged to take their seats if the seats are enough.
     

     

    Output
    For each test case, output the minimum number of chair that TIANKENG needs to prepare.
     

     

    Sample Input
    2
    2
    6 08:00 09:00
    5 08:59 09:59
    2
    6 08:00 09:00
    5 09:00 10:00
     

     

    Sample Output
    11
    6
     
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<algorithm>
     4 using namespace std;
     5 
     6 int main()
     7 {
     8     //freopen("in.txt","r",stdin);
     9     int sum[1550];
    10     int t,m,n,i,j,am,ah,lm,lh,ar,le,ans,num;
    11     scanf("%d",&t);
    12     while(t--)
    13     {
    14         memset(sum,0,sizeof(sum));
    15         scanf("%d",&n);
    16         for(i=0;i<n;i++)
    17         {
    18             scanf("%d %d:%d %d:%d",&num,&ah,&am,&lh,&lm);
    19             int ar=ah*60+am;
    20             int le=lh*60+lm-1;
    21             for(j=ar;j<=le;j++)
    22             sum[j]+=num;
    23         }
    24         ans=0;
    25         for(i=0;i<1550;i++)
    26         ans=max(ans,sum[i]);
    27         printf("%d
    ",ans);
    28     }
    29     return 0;
    30 }
  • 相关阅读:
    2014年寒假学习规划
    二十进制数的加法--【英雄会】
    使用IBM SVC构建vSphere存储间集群
    游戏服务器学习笔记 2———— 准备工作
    php判断正常访问和外部访问
    游戏服务器学习笔记 3———— firefly 的代码结构,逻辑
    数学基础知识 ——(1)高等数学
    动态内存与智能指针
    Numpy(4)—— 保存和导入文件
    Numpy(3)—— 线性代数相关函数
  • 原文地址:https://www.cnblogs.com/homura/p/4694298.html
Copyright © 2011-2022 走看看