POJ 2424 Flo's Restaurant 模拟

Flo's Restaurant

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2923		Accepted: 916

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

　　题目大意： 一家餐厅有a 个两人桌， b 个四人桌， c 个六人桌， 如果一组组人来吃饭， 如果对应的桌子（1， 2人对应两人桌， 以此类推）满的话就等， 如果发现等待超过30分钟就走，
假设每桌人都吃30分钟， 问这家餐馆最终能够招待多少人。

　　题目分析： 这道题归类是动态规划可我并没有发现， 用模拟做的， 一开始想的很简单， 知道等待队列里超过1人就不排队了（这是错的， 因为有多桌人在吃饭， 可能两桌吃完的间隔小于30分钟， 
如果说只有一桌那这种是对的）， 看来以后不能想当然啊！这道题的思路就是创建数组记录每桌的离开时间， 然后将排队的人与最早离开时间相比较， 这里有一个小小的trick就是
如果客人到达时发现所有桌都空， 则将客人到达的时间设定为最早桌进入的时间，代码注释中已经说明。

　　附上代码：
　　

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm> 
using namespace std;

int main() {
    int a, b, c;
    int x[101];
    int y[101];
    int z[101];
//    freopen( "in.txt", "r", stdin );
    while( cin >> a >> b >> c ) {
        if( !a && !b && !c ) break;
        memset( x, 0, sizeof( x ) );
        memset( y, 0, sizeof( y ) );
        memset( z, 0, sizeof( z ) );
        string s;
        int sum = 0;
        while( cin >> s && s != "#" ) {
            int n;
            int time = ( ( s[0] - '0' ) * 10 + s[1] - '0' - 8 ) * 60 + (s[3] - '0') * 10 + s[4] - '0';
            cin >> n;
            switch ( n ) {
                case 1:
                case 2: {
                    sort( x, x+a );   // x记录的是所有同一编号桌子的客人的出来时间， x[0]为最早桌出来的时间
                    if( x[0] - time <= 30 ) { // 等的时间不超过30分钟才排队
                        sum += n;
                        if( x[0] < time ) x[0] = time;   //如果客人到达时发现所有桌都空， 则将客人到达的时间设定为最早桌进入的时间
                        x[0] += 30;  // 最早桌出来的时间存入x[0]
                    }
                    break;
                }
                case 3:
                case 4: {
                    sort( y, y+b );
                    if( y[0] - time <= 30 ) {
                        sum += n;
                        if( y[0] < time ) y[0] = time;
                        y[0] += 30;
                    }
                    break;
                }
                case 5:
                case 6: {
                    sort( z, z+c );
                    if( z[0] - time <= 30 ) {
                        sum += n;
                        if( z[0] < time ) z[0] = time;
                        z[0] += 30;
                    }
                    break;
                }
                default:
                    break;
            }
        }
        cout << sum << endl;
    }
    return 0;
}

　　自己还是欠练啊， 效率还是不高啊！ 加油！