[hdu 1260] Tickets [DP]

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

我一直再找的方程中有三个状态，弄的我蒙蒙的，最后还是没有高出来，其实只用两个状态就行啦，简单的题，被我弄得复杂啦。[好菜啊] TAT
分析
我们定义 dp[i] 表示 从第一个人到第i个人的最小时间
对于每个人，他要么就是dp[i-1] 加上自己的单独时间 ，要么就是dp[i-2]加上自己和前一个一起买的时间，求个最小就行啦。
dp[i]= min(dp[i-1]+cost[i],dp[i-2]+tegother[i-1]) ;
代码

#include<cstdio>#include<iostream>#include<cstring>using namespace std;const int MAXN =2000+10;const int MAXM =1e5;const int inf  = 0x3f3f3f3f;int arr[MAXN];int brr[MAXN];int dp[MAXN];int main(){    int T;cin>>T;    while(T--){        int hh=8;int mm=0;int ss=0;        int k;scanf("%d",&k);        for(int i=1;i<=k;i++) scanf("%d",&arr[i]);        for(int i=1;i<=k-1;i++) scanf("%d",&brr[i]);                dp[1]=arr[1];dp[0]=0;            for(int i=2;i<=k;i++){                       dp[i]=min(dp[i-1]+arr[i],dp[i-2]+brr[i-1]);            }         int ans=dp[k];         ss+=ans;          mm+=ss/60;  ss=ss%60;         hh+=mm/60;  mm=mm%60;        printf("%02d:%02d:%02d ",hh,mm,ss);        if(hh>=12) puts("pm");        else puts("am");    }    return 0 ;}