hdu1260 Tickets（DP）

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本文介绍了一个关于电影票销售的算法问题，通过动态规划的方法求解售票员将票卖给特定数量顾客所需的最短时间。问题允许两种票务处理方式：单独售卖给一位顾客或同时售卖给两位相邻顾客。

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1260

Tickets

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2897 Accepted Submission(s): 1413

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 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

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

题意：求售票员把票卖给K个人所需的最短时间，票有2种方式售卖，一种是单独卖给一个顾客，另一种就是卖给相邻的两个顾客（_(:з」∠)_看到这里就觉得是dp了）

解题思路：首先，票有2种方式售卖，就猜想有2个转移方程。这时我们先设dp[i]为卖给前i个人（包含第i个人）的最短时间，很明显有2种状态可以转移到这个状态上。

①第i个人单独购买票。即dp[i-1] + s[i] ; （s[i]为第i个人单独购票所需的时间）

②第i和i-1一起买票。即dp[i-2] +d[i-1] ；（d[i]第i和第i-1一起买票所需的时间）

所以我们就可以很简单得出方程 dp[i] = min { dp[i-1] + s[i] , dp[i -2] +d[i-1]}

AC代码如下：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<set>
#include<cmath>
#include<vector>
#include<map>
using namespace std;
typedef long long ll;
const int maxn = 2050;
const int INF = 1e+9;
int d[maxn];
int s[maxn];
int dp[maxn];
char str[][10]={"pm","am"};
int main()
{
    int n;
    cin >> n ;
    while(n--)
    {
        int k1 ;
        scanf("%d",&k1);
        for(int i = 1 ; i <= k1 ; i ++)
            scanf("%d",&s[i]);
        int k2 = k1-1;
        for(int i = 1 ; i <= k2 ; i ++)
            scanf("%d",&d[i]);
        dp[0] = 0;
        for(int i = 1; i <= k1; i ++)
            dp[i] = INF;
        for(int i = 1 ; i <= k1; i ++)
        {
            dp[i] = min(dp[i],dp[i-1]+s[i]);
            if(i>=2)
                dp[i] =min(dp[i],dp[i-2]+d[i-1]);
        }
        int cnt1 = dp[k1]/3600;
        int cnt2 = (dp[k1]-cnt1*3600)/60;
        int cnt3 = dp[k1]%60;
        int HH,MM,SS,flag=1;
        HH = 8+cnt1 ;
        MM = cnt2;
        SS = cnt3;
        if(HH>=12)
        {
            HH-=12;
            flag = 0;
        }
        printf("%02d:%02d:%02d %s\n",HH,MM,SS,str[flag]);
    }

    return 0 ;
}