[ACM] HDU 4883 TIANKENG’s restaurant

TIANKENG’s restaurant

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 931    Accepted Submission(s): 412

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

 

226 08:00 09:005 08:59 09:5926 08:00 09:005 09:00 10:00

 

 

Sample Output

 

116

 

 

Source

 

BestCoder Round #2

 

解题思路：

题意为：有n组客人来吃饭，给出每组客人的人数及用餐开始时间，结束时间，格式为hh:mm；要求一组客人来的时候就必须给其安排位子
，问最少需要多少把椅子才能做到（一位客人需要一把椅子）.

time[i]，表示第i分钟有多少用餐的人，也就是需要多少把椅子，将开始时间，结束时间转化为分钟为单位的时间。
注意边界一组的结束和另一组的开始如果相同，则不需要额外的椅子，因此把每组的结束时间都-1. 对于每一组人，开始时间到结束时间
循环time[i]+=该组的人数。  最后再遍历time[i]数组，从中找到最大值即为该题的答案。

代码：

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>#include <stdlib.h>#include <cmath>#include <iomanip>#include <vector>#include <set>#include <map>#include <stack>#include <queue>#include <cctype>using namespace std;#define ll long longint n;int time[1442];int main(){    int t;    scanf("%d",&t);    while(t--)    {        memset(time,0,sizeof(time));        scanf("%d",&n);        int sh,sm;        int eh,em;        int cnt=0;        int ans=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&cnt);            scanf("%d:%d",&sh,&sm);            scanf("%d:%d",&eh,&em);            int s=sh*60+sm;            int e=eh*60+em;            e--;            for(int i=s;i<=e;i++)            {                time[i]+=cnt;            }        }        for(int i=0;i<1440;i++)        {            if(ans<time[i])                ans=time[i];        }        printf("%d\n",ans);    }    return 0;}