hd4883 TIANKENG’s restaurant

TIANKENG’s restaurant

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1626    Accepted Submission(s): 585

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

 

恩，题目大意就是说，饭店座位问题，给出每群人的人数以及到来和离开时间，问饭店总需要最少准备多少凳子，其实就是求区间最大覆盖问题。这里需要将时间转化为分钟进行储存标记，贪心思想也是，开始想不到转化，后来学长给讲了之后才懂。和南阳上一道旅馆的题相似，求最少房间数好像是，那天有空找一下贴出来。

#include<cstdio>#include<cstring>int flag[10010];int main(){int t,n,p,sh,sm,eh,em;scanf("%d",&t);while(t--){int maxi=0;memset(flag,0,sizeof(flag));scanf("%d",&n);while(n--){scanf("%d %d:%d %d:%d",&p,&sh,&sm,&eh,&em);for(int i=sh*60+sm;i<eh*60+em;++i){flag[i]+=p;if(flag[i]>maxi)maxi=flag[i];}}printf("%d\n",maxi);}return 0;}