• 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:
    UNP(2rd)第二卷源码编译
    A very hard mathematic problem
    并查集~
    N皇后问题 深搜dfs
    实践中的一点小问题
    java环境配置 HelloWorld
    精确时间计算方法
    C语言关于文件操作
    字典树应用~
    并查集应用2
  • 原文地址:https://www.cnblogs.com/ccccnzb/p/3875014.html
Copyright © 2020-2023  润新知