HDU 1260 tickets (motion gauge)

Tickets Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 924 accepted submission (s): 468


Problem descriptionjesus, 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 cocould he go back home as early as possible.
A good approach, cing the total time of tickets selling, is let adjacent people buy tickets together. as the restriction of the ticket seller machine, Joe can handle 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. cocould you so kind to tell poor Joe at what time cocould he go back home as early as possible? If so, I guess Joe wocould full of appreciation for your help.
 
Inputthere 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 (0 S <= SI <= 25 s) representing the time consumed to buy a ticket for each person;
(K-1) integer numbers (0 S <= di <= 50 s) representing the time needed for two adjacent people to buy two tickets together.
 
Outputfor every scenario, please tell Joe at what time cocould 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

2220 254018

 
Sample output

08:00:40 am08:00:08 am

 
Source Zhejiang University of Technology's fourth college Program Design Competition
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Solution:

A person can buy a ticket separately for a certain period of time, or two people can buy a ticket together, and a time is also given,

Give K individual separately bought time and K-1 two adjacent two people bought time together, ask the minimum time spent in total.

Map [] indicates the time a person bought separately, and V [] indicates the time when two people bought together.
The decision is to buy them together or separately;

Code: 0 ms

# Include <cstdio> # include <iostream> # include <cstring> # include <algorithm>Using namespaceSTD;# Define M 2050 # define INF 999999IntMap[M],V[M],DP[M]; Int main () {intI,N,M,T,J,K; While (CIN>T) {While (T--){CIN>N; (I=1;I<=N;I++)CIN>Map[I]; // The time when the ticket is bought separately. For (I=1;I<=N-1;I++)CIN>V[I]; // The time when the ticket is bought together. For (I=0;I<=N;I++)DP[I] = (I=0?0:INF);DP[1] =Map[1]; (I=2;I<=N;I++ ){DP[I] =Min(DP[I-1] +Map[I],DP[I-2] +V[I-1]); // Decision. } IntH=DP[N]/3600+8; // Converts the time. IntM=DP[N]/60%60; IntS=DP[N] %60;Printf("% 02d: % 02d: % 02d am \ n",H,M,S); // No two replicas are added to the first zero. } Return0;}