杭电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): 1801    Accepted Submission(s): 657

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

 

这个题大家都把他当成贪心了，其实不是的，这是一个区间覆盖和更新问题，把时间都换成分钟，更新区间求出最大值，要注意一点，6 08:00 09:005 09:00 10:00这个例子说明重叠的区间是不会算到上一个区间内的！

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int time[1500];int main(){int m,n,i,j,N,a,b,c,d,num;scanf("%d",&N);while(N--){scanf("%d",&m);memset(time,0,sizeof(time));for(i=0;i<m;i++){scanf("%d%d:%d%d:%d",&num,&a,&b,&c,&d);time[a*60+b]+=num;time[c*60+d]-=num;}int sum=0;for(i=1;i<=1440;i++){time[i]+=time[i-1];sum=max(time[i],sum);}printf("%d\n",sum);}return 0; }