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
题意：
Joe 想花费最少时间卖票给n个人，相邻两人可以单独买票也可以一起卖票，问最少时间
思路：
从第一个人往后，每个人买票都只有两种可能，一种是自己买，一种是和前面的人一起买，DP
1 #include <bits/stdc++.h>
2
3 using namespace std;
4
5 int main()
6 {
7 int t, n, i, a[2005], b[2005];
8 int dp[2005], h, m, s;
9 scanf("%d", &t);
10 while(t--)
11 {
12 scanf("%d", &n);
13 for(i=0;i<n;i++)
14 {
15 scanf("%d", &a[i]);
16 }
17 for(i=0;i<n-1;i++)
18 {
19 scanf("%d", &b[i]);
20 }
21
22 dp[0] = a[0]; //第0个人前面没人，只能自己买
23 dp[1] = min(a[0]+a[1], b[0]);
24 for(i=2;i<n;i++)
25 {
26 dp[i] = min(dp[i-1]+a[i], dp[i-2]+b[i-1]);
27 //自己买，前面i-1取最优，或者是和前面的人一起买，那么前面i-2个人取最优
28 }
29
30 h = 8;
31 m = 0;
32 s = 0;
33 s += dp[n-1];
34 m = m + s / 60;
35 s %= 60;
36 h = h + m / 60;
37 m %= 60;
38 if(h>12) printf("%02d:%02d:%02d pm\n", h-12, m, s);
39 else if(h==12) printf("%02d:%02d:%02d pm\n", h, m, s);
40 else printf("%02d:%02d:%02d am\n", h, m, s);
41
42 }
43 return 0;
44 }