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.
OutputFor 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 10:00Sample Output
11
6
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define maxn 2000
int t,n,a[maxn];
int main()
{
scanf("%d", &t);
while(t--)
{
memset(a, 0, sizeof(a));
scanf("%d", &n);
int num,hour1,minute1,hour2,minute2;
for(int i=0; i<=n-1; i++)
{
scanf("%d%d:%d%d:%d",&num,&hour1,&minute1,&hour2,&minute2);
int num1=hour1*60+minute1,num2=hour2*60+minute2;
for(int j=num1; j<=num2-1; j++)
{
a[j]+=num;
}
}
int ans = -1;
for(int i=0; i<=maxn-1; i++)
{
ans = max(ans,a[i]);
}
printf("%d\n", ans);
}
}