zoukankan      html  css  js  c++  java
  • POJ3186(KB12-O DP)

    Treats for the Cows

    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 5801   Accepted: 3003

    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

     
     1 //2017-04-06
     2 #include <iostream>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <algorithm>
     6 
     7 using namespace std;
     8 
     9 const int N = 2005;
    10 int v[N], n, dp[N][N];//dp[l][r]表示区间l~r间的最大收益
    11 //状态转移方程:dp[l][r] = max(dp[l+1][r]+day*v[l], dp[l][r-1]+day*v[r])
    12 
    13 int dfs(int l, int r, int day)
    14 {
    15     if(l > r)return 0;
    16     if(dp[l][r])return dp[l][r];
    17     if(l == r)return dp[l][r] = day*v[l];
    18     return dp[l][r] = max(dfs(l+1, r, day+1)+day*v[l], dfs(l, r-1, day+1)+day*v[r]);
    19 }
    20 
    21 int main()
    22 {
    23     while(scanf("%d", &n)!=EOF)
    24     {
    25         for(int i = 0; i < n; i++)
    26               scanf("%d", &v[i]);
    27         memset(dp, 0, sizeof(dp));
    28         int ans = dfs(0, n-1, 1);
    29         printf("%d
    ", ans);
    30     }
    31 
    32     return 0;
    33 }
  • 相关阅读:
    Ping
    boost::python开发环境搭建
    mingw和libcurl
    ssh远程执行命令使用明文密码
    netty源码阅读之UnpooledByteBufAllocator
    Direct ByteBuffer学习
    clions的使用
    netty中的PlatformDependent
    STL之priority_queue(优先队列)
    c++线程调用python
  • 原文地址:https://www.cnblogs.com/Penn000/p/6674530.html
Copyright © 2011-2022 走看看