HDU1260 Tickets DP

Tickets

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

题解

简单的DP题，只是题目一开始有点难懂
状态转移方程是：dp[i] = min(dp[i-1]+单独买花的时间, dp[i-2]+和前面那个人一起买花的时间
初始状态是dp[1] = 第一个人单独买花的时间

代码

#include
#include
#include
#include
#include
using namespace std;

int dp[2001];
int s[2001];
int d[2001];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
       int n;
       scanf("%d",&n);
       for (int i = 1;i<=n;i++)
        scanf("%d",&s[i]);
       for (int i = 2;i<=n;i++)
        scanf("%d",&d[i]);
       dp[1] = s[1];
       for (int i = 2;i<=n;i++)
        dp[i] = min(s[i]+dp[i-1],dp[i-2]+d[i]);

       int hh = dp[n]/ 3600;
       int mm = dp[n]%3600/60;
       int ss = dp[n]%60;
       printf("%02d:%02d:%02d am\n",(8+hh)%24,mm,ss);
    }
}