HDU 1103 Restaurant

Flo's Restaurant

                                                                    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768
K (Java/Others)

                                                                                                       Total Submission(s): 1463    Accepted Submission(s): 440


Problem Description

Sick and tired of pushing paper in the dreary bleary-eyed world of finance, Flo ditched her desk job and built her own restaurant.

In the small restaurant, there are several two-seat tables, four-seat tables and six-seat tables. A single diner or a group of two diners should be arranged to a two-seat table, a group of three or four diners should be arranged to a four-seat table, and a
group of five or six diners should be arranged to a six-seat table.

Flo’s restaurant serves delicious food, and many people like to eat here. Every day when lunch time comes, the restaurant is usually full of diners. If there is no suitable table for a new coming group of diners, they have to wait until some suitable table
is free and there isn’t an earlier arrival group waiting for the same kind of tables. Kind Flo will tell them how long they will get their seat, and if it’s longer than half an hour, they will leave for another restaurant.

Now given the list of coming diners in a day, please calculate how many diners take their food in Flo’s restaurant. You may assume it takes half an hour for every diner from taking a seat to leaving the restaurant.

 

Input

There are several test cases. The first line of each case contains there positive integers separated by blanks, A, B and C (A, B, C >0, A + B + C <= 100), which are the number of two-seat tables, the number of four-seat tables and the number of six-seat tables
respectively. From the second line, there is a list of coming groups of diners, each line of which contains two integers, T and N (0 < N <= 6), representing the arrival time and the number of diners of each group. The arrival time T is denoted by HH:MM, and
fixed between 08:00 and 22:00 (the restaurant closes at 23:00). The list is sorted by the arrival time of each group in an ascending order, and you may assume that no groups arrive at the same time. Each test case is ended by a line of “#”.

A test case with A = B = C = 0 ends the input, and should not be processed.

 

Output

For each test case, you should output an integer, the total number of diners who take their food in Flo’s restaurant, in a separated line.

 

Sample Input

1 1 1
10:40 1
10:50 2
11:00 4
#
1 1 1
10:40 1
10:50 2
11:00 2
#
1 2 1
10:30 1
10:40 3
10:50 2
11:00 1
11:20 5
#
0 0 0

 

Sample Output

7
3
12

 

Author

Alcyone

 

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题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1103

分析：

优先队列Table[ i ] ,下标表示桌子类型编号1、2、3，优先队列里保存同类型所有桌子上的

客人用餐结束时间（越小越优先）

每当有顾客来的时候，判断是否有桌子可以坐下来，并更新队列

代码：

#include <cstdio>
#include <queue>
#include <functional>
using namespace std;
//3种桌子分别用下标为1、2、3的优先队列表示
priority_queue<int, vector<int>, greater<int> > Table[5];

int main()
{
int a, b, c;
while (~scanf("%d%d%d", &a, &b, &c))
{
if (a + b + c == 0)
break;
for (int i = 1; i <= 3; i++)
{
while (!Table[i].empty())
Table[i].pop();
}

for (int i = 1; i <= a; i++) //将每种桌子加入队列
Table[1].push(0);
for (int i = 1; i <= b; i++)
Table[2].push(0);
for (int i = 1; i <= c; i++)
Table[3].push(0);

char Time[10];
int num, cnt = 0;
while (1)
{
scanf("%s", Time);
if (Time[0] == '#')
break;
scanf("%d", &num);
int hh = 10 * (Time[0] - '0') + Time[1] - '0';
int mm = 10 * (Time[3] - '0') + Time[4] - '0';
int Merge = hh * 60 + mm; //将时间转换为分钟表示
int index = (num + 1)/ 2; //由顾客数目得到桌子种类的编号
if (Merge + 30 >= Table[index].top()) //判断是否能坐下吃美味
{
cnt += num;
//此处注意，求当前顾客的开始吃饭时间，
//应为上一波顾客的结束时间与当前顾客的开始时间中的较大者
int Max = max(Table[index].top(), Merge);
Table[index].push(Max + 30); //将当前顾客的结束时间加入队列
Table[index].pop();
}
}
printf("%d\n", cnt);
}
return 0;
}