Tickets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1916    Accepted Submission(s): 929

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

题意：有N个人排队买电影票，已知每个人买票需要的时间和两个人（连续的两人）一起买票需要的时间。问最少需要多少时间才能卖完所有票。

思路：用dp[ i ]表示前 i 个人买票需要的最短时间。man[ i ]表示第i个人单独买票需要的时间，two[ i ]表示第i-1和第i个人一起买票需要的时间。

则状态转移方程——dp[ i ] = min(dp[ i -1] + man[ i ], dp[ i - 2] + two[ i ] )。

AC代码：

#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <map>
#include <string>
#include <cstdlib>
#include <algorithm>
#define MAXN 2000+10
#define MAXM 60000+10
#define INF 1000000
#define eps 1e-8
#define LL long long
using namespace std;
int N;
int man[MAXN];
int two[MAXN];
int dp[MAXN];
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        for(int i = 1; i <= N; i++)
            scanf("%d", &man[i]);
        for(int i = 2; i <= N; i++)
            scanf("%d", &two[i]);
        memset(dp, 0, sizeof(dp));
        dp[1] = man[1];
        for(int i = 2; i <= N; i++)
            dp[i] = min(dp[i-1] + man[i], dp[i-2] + two[i]);
        int t = 28800 + dp[N];
        int hou = t / 3600;
        int mie = t % 3600 / 60;
        int sec = (t % 3600) % 60;
        if(t >= 43200)
        {
            if(hou > 12)
                hou -= 12;
            printf("%02d:%02d:%02d pm\n", hou, mie, sec);
        }
        else
            printf("%02d:%02d:%02d am\n", hou, mie, sec);
    }
    return 0;
}

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