HDU：4883 TIANKENG’s restaurant（区间计数）

TIANKENG’s restaurant

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 2559    Accepted Submission(s): 951


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

BestCoder Round #2

 

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题目大意：

给你t批客人，每批有n个客人，给出每批客人来的时间段，并且每个来的客人都得拥有一个凳子，问这个饭店至少需要准备多少个凳子。

解题思路：

把时间变成数组区间，给区间加人数，然后求出人数最多的那个点对应的人数。

代码如下：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
bool cmp(int a,int b)
{
return a>b;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int group;
int a[1450];
memset(a,0,sizeof(a));
scanf("%d",&group);
for(int i=0;i<group;i++)
{
int people,t1,t2,t3,t4;
scanf("%d %d:%d %d:%d",&people,&t1,&t2,&t3,&t4);
int s=t1*60+t2;
int e=t3*60+t4;
for(int j=s+1;j<=e;j++)
{
a[j]=a[j]+people;
}
}
sort(a,a+1450,cmp);
printf("%d\n",a[0]);
}
return 0;
}