POJ 3189 Treats for the cows (two DP methods resolved)

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4264		Accepted: 2155

Description

FJ have purchased N (1 <= n <=) 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 is interesting for many reasons:

• The treats is numbered 1..N and stored sequentially in single file in a long box, which is open at both ends. On any day, the FJ can retrieve one treat from either end of the stash.
• Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
• The treats is not uniform:some is better and has higher intrinsic value. Treat I has value V (i) (1 <= v (i) <= 1000).
• Cows pay more for treats that has aged longer:a cow would pay V (i) *a for a treat of age a.

Given The values V (i) of each of the treats lined up and order of the index I in their box, what is the greatest value FJ C A receive for them if he orders their sale optimally?

The first treat are sold on Day 1 and have 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.

Method One

According to test instructions Direct Dp,dp[i][j][2],i is the first day, J is there J, 0 is taken from the front, 1 is taken from the back. Then it can be launched according to Dp[i-1][j][0],dp[i-1][j][1] dp[i][j][1], according to Dp[i-1][j-1][0],dp[i-1][j-1][1] launch dp[i][j][0].

Code:

#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;}

Method two (interval DP)

The dp[i][j],i represents the interval starting point, J represents the interval end point, and LG represents the interval length.

Code:

#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;//interval length is 0, only one number, is the last number 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));                Here is a push forward from the beginning of the last team, I~j can be launched from I+1~j and I~j-1            }        }        printf ("%d\n", Dp[1][n]);}    }

POJ 3189 Treats for the cows (two DP methods resolved)