hduTickes("简单"的dp)

想退役了~

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

2 2 20 25 40 1 8

 

Sample Output

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

问题分析:线性dp,一开始想的十分的复杂,想了三种转移状态还有端点处的特判,后来才发现其实只有两种.

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m;
int time[2005],total1[1000];
int a[2005],b[2005];
int fen,miao,shi,total;
int s[2],f[2],mi[2];
int main()
{
    cin>>n;
    while(n--)
    {
        cin>>m;
        if(m!=1)
        {
            for(int i=1; i<=m; i++)
                cin>>a[i];
            for(int i=2; i<=m; i++)
            {
                cin>>time[i];
            }
            total1[0]=0;
            total1[1]=a[1];
            for(int i=2;i<=m;i++)
            {
                total1[i]=min(total1[i-1]+a[i],total1[i-2]+time[i]);
            }
            total=total1[m];
        }
        if(m==1)
        {
            cin>>a[m];
            total+=a[m];
        }
        miao=total%60;
        fen=(total-miao)/60;
        shi=fen/60;
        fen=fen%60;
        shi+=8;
        f[0]=fen/10;
        f[1]=fen%10;
        s[0]=shi/10;
        s[1]=shi%10;
        mi[0]=miao/10;
        mi[1]=miao%10;
        if(shi<12)
            cout<<s[0]<<s[1]<<':'<<f[0]<<f[1]<<':'<<mi[0]<<mi[1]<<' '<<"am"<<endl;
        else
            cout<<s[0]<<s[1]<<':'<<f[0]<<f[1]<<':'<<mi[0]<<mi[1]<<' '<<"pm"<<endl;
        total=0;
        shi=0;
        miao=0;
        fen=0;
    }
    return 0;
}