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:
    一个简单的空间配置器
    【转】c++中placement new操作符
    类的operator new与operator delete的重载【转】
    STL中常用的c++语法
    java编程思想-多态
    java编程思想-复用类(2)
    java编程思想-复用类
    import与require的区别
    gulp插件
    gulp-gulpfile.js语法说明
  • 原文地址:https://www.cnblogs.com/homura/p/4694298.html
Copyright © 2011-2022 走看看