杭电oj--Tickets(dp)

最新推荐文章于 2024-08-03 14:51:53 发布

转载最新推荐文章于 2024-08-03 14:51:53 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/soTired/p/4737650.html

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#java

本文介绍了一个关于电影票销售的问题，旨在通过优化算法减少售票总时间。问题设定为：有N个场景，每个场景包含一组观众及其单独购票时间和两人一起购票的时间。目标是通过动态规划算法找出使售票员最早下班的方案。

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Tickets

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1941 Accepted Submission(s): 938

Problem Description

Jesus, what a great movie! Thousands of people are rushing to the cinema. However, this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible.
A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together. As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time.
Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her. Could you so kind to tell poor Joe at what time could he go back home as early as possible? If so, I guess Joe would full of appreciation for your help.

Input

There are N(1<=N<=10) different scenarios, each scenario consists of 3 lines:
1) An integer K(1<=K<=2000) representing the total number of people;
2) K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;
3) (K-1) integer numbers(0s<=Di<=50s) representing the time needed for two adjacent(相近的) people to buy two tickets together.

Output

For every scenario, please tell Joe at what time could he go back home as early as possible. Every day Joe started his work at 08:00:00 am. The format of time is HH:MM:SS am|pm.

Sample Input

2 2 20 25 40 1 8

Sample Output

08:00:40 am 08:00:08 am

Source

浙江工业大学第四届大学生程序设计竞赛

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感觉dp真强大, dp[i]=min(dp[i-1]+num1[i] , dp[i-2]+num2[i-1]). 通过循环来比较求的最优解。 dp[i-1]+num1[i]每次+一个未买票的人所耗时间， dp[i-2]+num2[i-1]为两人共同买票（两次）所用时间。求最小值。

-------- 正午据说是pm , 说法存在争议。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #define min(a, b) a<b?a:b 
 5 using namespace std;
 6 int main()
 7 {
 8     int t;
 9     int num1[2020], num2[2020], dp[2020];
10     scanf("%d", &t);
11     while(t--)
12     {
13         int m;
14         scanf("%d", &m);
15         for(int i = 1; i <= m; i++)
16             scanf("%d", &num1[i]);
17         for(int i = 1; i <= m-1; i++)
18             scanf("%d", &num2[i]);
19         memset(dp, 0, sizeof(dp));
20         dp[1] = num1[1];
21         for(int i = 2; i <= m; i++)
22             dp[i] = min(dp[i-1] + num1[i], dp[i-2] + num2[i-1]);
23         int b, c, d;
24         b = dp[m] % 60;
25         c = dp[m] / 60 % 60;
26         d = dp[m] /3600;
27         d += 8;
28         char c1, c2;
29         if(d < 12)
30         {
31             c1 = 'a';
32             c2 = 'm';
33         }
34         else
35         {
36             d %= 12;
37             c1 = 'p';
38             c2 = 'm';
39         }
40         printf("%02d:%02d:%02d %c%c\n", d, c, b, c1, c2);
41     }
42     return 0;
43 }