HDU 1260 Tickets 简单dp

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本文介绍了一道关于电影票销售的算法题目，旨在通过优化售票流程来减少顾客等待时间。文章详细阐述了问题背景及输入输出要求，并给出了使用动态规划解决该问题的代码示例。

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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): 4121 Accepted Submission(s): 2074

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

题意：每个人都有一个买票时间，并且又给出相邻两个人的买票时间，求最少多长时间能买完票。

思路：用dp[i]存前i个人的最小花费时间。

AC代码：

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<string>
#define LL long long
#define eps 1e-8
using namespace std;
const int mod = 1e7+7;
const int INF = 1e8;
const int inf = 0x3f3f3f3f;
const int maxx = 100100;
const int N = 2050;
int a[N];
int b[N];
int dp[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        for(int i=2; i<=n; i++)
            scanf("%d",&b[i]);
        memset(dp,0,sizeof(dp));
        dp[1]=a[1];
        for(int i=2; i<=n; i++)
            dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);
        int shi=8,fen=0,miao=0;
        string c="am";
        miao=dp[n];
        fen+=miao/60;
        miao%=60;
        shi+=fen/60;
        fen%=60;
        if(shi>12)
        {
            shi-=12;
            c="pm";
        }
        printf("%02d:%02d:%02d ",shi,fen,miao);
        cout<<c<<endl;
    }
}