POJ 3189 Treats for the Cows(两种DP方法解决)

Treats for the Cows

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4264 Accepted: 2155

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

513152

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.

方法一

       按照题意直接dp，dp[i][j][2],i是第i天，j是有j个,0是从前面取，1是从后面取。接着就可以根据dp[i-1][j][0],dp[i-1][j][1]推出dp[i][j][1],根据dp[i-1][j-1][0],dp[i-1][j-1][1]推出dp[i][j][0]。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=2000+100;int a[maxn];int dp[maxn][maxn][2];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)        scanf("%d",&a[i]);        memset(dp,0,sizeof(dp));        int ans=max(a[1],a[n]);        dp[1][1][0]=a[1];        dp[1][0][1]=a[n];        for(int i=2;i<=n;i++)        {            for(int j=i;j>=0;j--)            {                if((dp[i-1][j-1][0]||dp[i-1][j-1][1])&&j>0)                dp[i][j][0]=max(dp[i-1][j-1][1],dp[i-1][j-1][0])+a[j]*i;                if(dp[i-1][j][0]||dp[i-1][j][1])                dp[i][j][1]=max(dp[i-1][j][0],dp[i-1][j][1])+a[n-i+j+1]*i;                ans=max(ans,dp[i][j][0]);                ans=max(ans,dp[i][j][1]);            }        }        printf("%d\n",ans);    }    return 0;}

方法二(区间DP）

    dp[i][j],i表示区间起始点，j表示区间结束点，lg表示区间长度。

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int maxn=2000+200;int a[maxn];int dp[maxn][maxn];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);        }        for(int i=1;i<=n;i++)        dp[i][i]=a[i]*n;//区间长度为0，只有一个数，是最后出的数        for(int lg=1;lg<n;lg++)        {            for(int i=1;i<=n;i++)            {                int j=i+lg;                dp[i][j]=max(dp[i+1][j]+a[i]*(n-lg),dp[i][j-1]+a[j]*(n-lg));                //这里是从最后出队的开始往前推，i~j可以从i+1~j和i~j-1推出            }        }        printf("%d\n",dp[1][n]);    }}