zoukankan      html  css  js  c++  java
  • hdu 4883 思维题

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=4883

    TIANKENG’s restaurant

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 566    Accepted Submission(s): 267


    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
     
    Source

     ---------------------------------------------------------------------

     1 #include <stdio.h>
     2 #include <stdlib.h>
     3 #include <string.h>
     4 
     5 int main()
     6 {
     7     int str[1445];
     8     int n,m,i,j,k,t,num,hh1,hh2,mm1,mm2;
     9     scanf("%d",&n);
    10     while(n--)
    11     {
    12         memset(str,0,sizeof(str));
    13         scanf("%d",&m);
    14         for(i=0;i<m;i++)
    15         {
    16             scanf("%d %d:%d %d:%d",&num,&hh1,&mm1,&hh2,&mm2);
    17             hh1=hh1*60+mm1;
    18             hh2=hh2*60+mm2;
    19             str[hh1]+=num;
    20             str[hh2]-=num;//printf("%d%d^^",str[hh1],str[hh2]);
    21         }
    22         int maxx=0,cas=0;
    23         for(i=0;i<=1440;i++)
    24         {
    25             cas+=str[i];
    26             if(cas>maxx)
    27             {
    28                 maxx=cas;
    29             }
    30         }
    31         printf("%d
    ",maxx);
    32     }
    33     return 0;
    34 }
    View Code
  • 相关阅读:
    asp.net将word转成HTML
    asp.net 控件位置form验证
    javascript 正则表达式
    .net framework处理xml
    XMLHTTP与XMLDOC使用介绍(上)
    一个女程序员的征婚信息的select语句
    解决连接MySql4.0数据库出现的"character_set_results"问题
    javascript与cs代码互访
    杜绝<script src="js/tool.js" type="text/javascript"/>这样写的习惯
    总结:常用的正则表达式集合
  • 原文地址:https://www.cnblogs.com/ccccnzb/p/3875014.html
Copyright © 2011-2022 走看看