【贪心】hdoj4883 TIANKENG’s restaurant

最新推荐文章于 2022-04-16 17:04:49 发布

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最新推荐文章于 2022-04-16 17:04:49 发布

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分类专栏： ACM算法文章标签：贪心

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本文介绍了一个算法案例，通过计算不同时段顾客数量来确定餐厅所需的最小座位数，确保每位顾客到店都有座位可用。

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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

分析：假如来了四组客人，其时间段如下图所示，记录上起始时间和终止时间，在这段时间内所有客人的人数都要加上，最后找出最大的数组即可。

代码如下：

#include<stdio.h>
#include<string.h>
#define N 1440
int s[N];
int main()
{int T;
scanf("%d",&T);
while(T--)
{int n,i,j,h1,m1,h2,m2,m;
scanf("%d",&n);
memset(s,0,sizeof(s));
for(i=1;i<=n;i++)
  {scanf("%d %d:%d %d:%d",&m,&h1,&m1,&h2,&m2);

   for(j=h1*60+m1;j<h2*60+m2;j++)
   s[j]+=m;
  }
int max=0;
for(j=0;j<1440;j++)
if(s[j]>max) {max=s[j];}
printf("%d\n",max);
}
return 0;
}