【杭电oj】4883 - TIANKENG’s restaurant（区间覆盖）

TIANKENG’s restaurant

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2292    Accepted Submission(s): 835

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

在voj上提交的是从tst到ten-1各加tn的代码，可以AC的，但是在杭电上回超时。后来只好用区间更新的方法，顺利AC了。

代码如下：

#include <cstdio>#include <cstring>int max (int a,int b){if (a > b)return a;return b;}int main(){int u;int n;int time[1500];int ans;scanf ("%d",&u);while (u--){ans = -1;memset (time,0,sizeof (time));scanf ("%d",&n);for (int i = 0 ; i < n ; i++){int tn,tst,ten;scanf ("%d",&tn);int t1,t2,t3,t4;scanf ("%d:%d %d:%d",&t1,&t2,&t3,&t4);tst = t1 * 60 + t2;ten = t3 * 60 + t4;time[tst] += tn;time[ten] -= tn;//区间上限不做更新 }for (int j = 1 ; j <= 1400 ; j++)//j从2开始就WA，看来是00:00也算 {time[j] += time[j-1];ans = max (ans,time[j]);}printf ("%d\n",ans);}return 0;}