【杭电 4883】TIANKENG’s restaurant
来源:互联网 发布:网络店铺运营 编辑:程序博客网 时间:2024/04/30 22:14
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
记录下每分钟的人数,取人数最多的那分钟;
代码 :
#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;int a[100000+11]={0};int main(){ int t; scanf("%d",&t); while(t--) { int n,i,k,h1,h2,m1,m2,sum1,sum2,j,sum=0; scanf("%d",&n); memset(a,0,sizeof(a)); for(i=0;i<n;i++) { scanf("%d %d:%d %d:%d",&k,&h1,&m1,&h2,&m2); sum1=60*h1+m1; sum2=60*h2+m2; for(j=sum1;j<sum2;j++) { a[j]+=k; if(sum<a[j]) sum=a[j]; } } printf("%d\n",sum); } return 0;}
- HDU杭电4883 TIANKENG’s restaurant
- 杭电4883TIANKENG’s restaurant
- 【杭电】[4883]TIANKENG’s restaurant
- 【杭电4883】TIANKENG’s restaurant
- 杭电-4883 TIANKENG’s restaurant
- 【杭电4883】TIANKENG’s restaurant
- 【杭电 4883】TIANKENG’s restaurant
- 杭电4883-TIANKENG’s restaurant(区间覆盖)
- 【杭电oj】4883 - TIANKENG’s restaurant(区间覆盖)
- 4883 TIANKENG’s restaurant
- 杭电4883 TIANKENG’s restaurant(小板凳的问题)
- HDOJ 4883 TIANKENG’s restaurant
- HDU 4883 TIANKENG’s restaurant
- hdu 4883 TIANKENG’s restaurant
- HDU 4883 TIANKENG’s restaurant
- hdu 4883 TIANKENG’s restaurant
- hdu 4883 TIANKENG’s restaurant
- hdoj 4883 TIANKENG’s restaurant
- 单链表
- Ubuntu16.04 install CUDA8.0, CuDNN5.05,Pycuda
- Codeforces 698A: Vacations(贪心)
- 利用cookie显示上次浏览的时间
- CodeForces 610APasha and Stick (拆数字组成长方形)
- 【杭电 4883】TIANKENG’s restaurant
- JS正则验证手机号
- <a>标签的href和onclick属性
- java String.split()函数的用法分析
- hdu 3440 House Man(差分约束)
- View的事件分发机制
- redis配置安装和使用
- POJ 1753 Flip Game 暴力搜索(dfs加枚举)
- mybatis批量更新和插入