hdu4883 & BestCoder Round #2 TIANKENG’s restaurant（暴力）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4883

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 

题意：一家饭店在不同的时间段会有一些客人来就餐，给出不同时间段来就餐的客人的人数和到来的时间和离开的时间，求最少需要多少椅子才能满足需求！

其实就是一个求区间最大相交的问题；

这题暴力有点费时间（却没有超时），但是这却是最直接和最容易想到的办法！当然也有线段树版的做法，读者可自行百度！

代码如下：

#include <cstdio>#include <cstring>int main(){    int t, n;    int i, j, num;    int s1, s2, s3, s4;    int c[1517];int be, en;    scanf("%d",&t);    while(t--)    {int max = 0;        scanf("%d",&n);memset(c,0,sizeof(c));        for(i = 1; i <= n; i++)        {            scanf("%d",&num);            scanf("%d:%d %d:%d",&s1,&s2,&s3,&s4);            be = s1*60+s2;            en = s3*60+s4;for(j = be; j < en; j++){c[j]+=num;}        }        for(i = 1; i <= 1440; i++)        {            if(c[i] > max)                max = c[i];        }        printf("%d\n",max);    }    return 0;}