hdu4883 TIANKENG’s restaurant

TIANKENG’s restaurant

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3069 Accepted Submission(s): 1134

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
比赛的时候wa了13次，一开始用贪心，后来才看清题，但是用的方法一直TE，并不是会场安排那样的贪心，而是另一种，我也不知道什么类型。
最后赛后看别人题解，就是化为分钟，在一天之中每个时间都要计算这点来了多少人或走了多少人，然后用sum统计，最大的sum给ans；

#include<stdio.h>
#include<string.h>
const int maxn=1445;
int b[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(b,0,sizeof(b));
        int ans=0,sum=0;
        int n;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            int a,x1,y1,x2,y2;
            scanf("%d",&a);
            scanf("%d:%d",&x1,&y1);
            b[x1*60+y1]+=a;
            scanf("%d:%d",&x2,&y2);
            b[x2*60+y2]-=a;
        }
        for(int i=0;i<=1440;i++)
        {
            sum+=b[i];
            if(ans<sum)
                ans=sum;
        }
        printf("%d
",ans);
    }
    return 0;
}