HDU 1260 Tickets dp

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本文介绍了一种通过优化购票流程来减少售票总时间的方法。针对单一购买与相邻两人同时购买的情况，采用动态规划策略实现最小化售票时间的目标，并提供AC代码实现。

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Tickets

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4773 Accepted Submission(s): 2477

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

Source
浙江工业大学第四届大学生程序设计竞赛

Recommend
JGShining

优化之后，只剩两种状态，一种是单独购买，另一种就是和前面的人一起买，得到两个状态转移方程.

dp[i][0]=min(dp[i-1][0],dp[i-1][1])+s[i](单独购买)

dp[i][1]=min(dp[i-2][0],dp[i-2][1])+d[i];（和前面的人一起买）

ac代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <stack>
#include <algorithm>
#define inf 0x3f3f3f3f
using namespace std;

int n,k,s[2005],d[2005];
int dp[2005][2];
int main(){
	
	
	scanf("%d",&n);
	while(n--){
		scanf("%d",&k);
		for(int i=1;i<=k;i++){
			scanf("%d",&s[i]);
		}
		for(int i=2;i<=k;i++){
			scanf("%d",&d[i]);
		}
		d[1]=s[1];
		memset(dp,0,sizeof(dp));
		//dp[0][0]=dp[0][1]=0;
		for(int i=1;i<=k;i++){
			dp[i][0]=min(dp[i-1][0],dp[i-1][1])+s[i];
			if(i > 1){
				dp[i][1]=min(dp[i-2][0],dp[i-2][1])+d[i];
			}
			else
				dp[i][1]=s[i];
		}
		int ans=min(dp[k][0],dp[k][1]);
		int h=ans/3600;
		ans%=3600;
		int m=ans/60;
		ans%=60;
		int miao=ans;
		if(8 + h > 12){
			h=(8+h)%12;
			printf("%02d:%02d:%02d pm\n",h,m,miao);
		}
		else{
			//h=(8+h)%12;
			printf("%02d:%02d:%02d am\n",8+h,m,miao);
		}
	}
	
	
	return 0;
}