zoukankan      html  css  js  c++  java
  • POJ 3186 Treats for the Cows(区间dp)

    传送门:

    http://poj.org/problem?id=3186

    Treats for the Cows
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 7672   Accepted: 4059

    Description

    FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of milk. FJ sells one treat per day and wants to maximize the money he receives over a given period time.

    The treats are interesting for many reasons:
    • The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats.
    • Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
    • The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
    • Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.
    Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally?

    The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.

    Input

    Line 1: A single integer, N

    Lines 2..N+1: Line i+1 contains the value of treat v(i)

    Output

    Line 1: The maximum revenue FJ can achieve by selling the treats

    Sample Input

    5
    1
    3
    1
    5
    2

    Sample Output

    43

    Hint

    Explanation of the sample:

    Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).

    FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.

    Source

     
    分析:
    题意:给出的一系列的数字,可以看成一个双向队列,每次只能从队首或者队尾出队,第n个出队就拿这个数乘以n,最后将和加起来,求最大和
     
    思路:由里向外逆推区间
     
    dp[i][j]表示左边取了i个数,右边取了j个数
    dp[i][j] = max(dp[i-1][j] + a[i]* (i+j), dp[i][j-1] + a[n-j+1]*(i+j));
    注意当ij为0的边界判断即可。
     
    code:
    #include<stdio.h>
    #include<string.h>
    #include<memory>
    using namespace std;
    #define max_v 2005
    int a[max_v];
    int dp[max_v][max_v];
    int main()
    {
        /*
        dp[i][j]表示左边取了i个数,右边取了j个数
        故
        dp[i][j] = max(dp[i-1][j] + a[i]* (i+j), dp[i][j-1] + a[n-j+1]*(i+j));
        注意当ij为0的边界判断即可。
        */
        int n;
        while(~scanf("%d",&n))
        {
            for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
            }
            memset(dp,0,sizeof(dp));
            int ans=0;
            for(int i=0;i<=n;i++)
            {
                for(int j=0;j+i<=n;j++)
                {
                    if(i==0&&j==0)
                        dp[i][j]=0;
                    else if(i==0)
                        dp[i][j]=dp[i][j-1]+a[n-j+1]*j;
                    else  if(j==0)
                        dp[i][j]=dp[i-1][j]+a[i]*i;
                    else
                        dp[i][j]=max(dp[i-1][j]+a[i]*(i+j),dp[i][j-1]+a[n-j+1]*(i+j));
                }
               ans=max(ans,dp[i][n-i]);
            }
            printf("%d
    ",ans);
        }
        return 0;
    }
     
     
  • 相关阅读:
    userdir 希望用户能够以http://X.X.X.X/~username 方式来访问自己的网页
    var_export() 函数的使用
    mb_detect_encoding — 检测字符的编码
    详解PHP fsockopen的使用方法
    jQuery 返回顶部
    Mysql函数
    sql where 1=1和 0=1 的作用
    Numpy基础学习(三)
    Numpy 中的矩阵
    Numpy数组的全通用函数
  • 原文地址:https://www.cnblogs.com/yinbiao/p/9348004.html
Copyright © 2011-2022 走看看