poj 3186 Treats for the Cows(滚动DP OR 记忆化搜索)

O - Treats for the Cows

 POJ - 3186 

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.

题目大意：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

状态：dp[i][j]表示第i次移出j位置所得到的最大值

状态转移方程：dp[i][j] = max(dp[i-1][j-1]+arr[j]*i,dp[i-1][j]+arr[N-(i-j)+1]*i)

除了背包外，第一次完全自己独立思考的一道DP题目；

#include <iostream>#include <stdio.h>#include <cstring>using namespace std;#define MAXN 2005#define INF 0x3f3f3f3fint arr[MAXN];int dp[MAXN][MAXN];///dp[i][j]表示第i次移出j位置所得到的最大值int main(){    int N;    while(~scanf("%d",&N)){        for(int i=1;i<=N;++i)            scanf("%d",&arr[i]);        memset(dp[1],0,sizeof(dp[1]));        dp[1][1]=1*arr[1],dp[1][0]=1*arr[N];        for(int i=2;i<=N;i++){  ///第i次移出            //dp[i][j] = -INF ;            for(int j=0;j<=i;j++){                if(j==0) dp[i][j] = dp[i-1][j]+arr[N-(i-j)+1]*i;                else if(j==i) dp[i][j] = dp[i-1][j-1]+arr[i]*i;                else dp[i][j] = max(dp[i-1][j-1]+arr[j]*i,dp[i-1][j]+arr[N-(i-j)+1]*i) ;            }        }        int mmax=-INF;        for(int i=0;i<N;i++)            mmax=max(mmax,dp[N][i]);        printf("%d\n",mmax);    }    return 0;}

自己闲的*疼尝试了一把滚动数组，突然下面的代码还是会蒙圈的。。。

#include <iostream>#include <stdio.h>#include <cstring>using namespace std;#define MAXN 2005#define INF 0x3f3f3f3fint arr[MAXN];int dp[3][MAXN];///dp[i][j]表示第i次移出j位置所得到的最大值int main(){    int N;    while(~scanf("%d",&N)){        for(int i=1;i<=N;++i)            scanf("%d",&arr[i]);        memset(dp[1],0,sizeof(dp[1]));        dp[1][1]=1*arr[1],dp[1][0]=1*arr[N];        int nw=1,old=2;        for(int i=2,j;i<=N;i++){  ///第i次移出            for(j=0,swap(nw,old);j<=i;j++){                if(j==0) dp[nw][j] = dp[old][j]+arr[N-(i-j)+1]*i;                else if(j==i) dp[nw][j] = dp[old][j-1]+arr[i]*i;                else dp[nw][j] = max(dp[old][j-1]+arr[j]*i,dp[old][j]+arr[N-(i-j)+1]*i) ;            }        }        int mmax=-INF;        int tmp = (N%2==0 ? 2 :1);        for(int i=0;i<N;i++)            mmax=max(mmax,dp[tmp][i]);        printf("%d\n",mmax);    }    return 0;}

其实最简单的还是网上的

由于每次要么从头取，要么从尾取，于是状态转移方程为：

dp[i][j]=max(dp[i-1][j]+v[i]*(i+j),dp[i][j-1]+v[n-j+1]*(i+j));

所以DP的题目还是在开始的时候状态的选取还是很重要的，这直接影响到了后来的状态转移方程的选取！！！谨记

#include<stdio.h>#include<algorithm>using namespace std;int dp[2005][2005];int v[2005];int max(int a,int b){if(a>b)return a;elsereturn b;}int main(){int i,j,n;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;i++)scanf("%d",&v[i]);memset(dp,0,sizeof(dp));for(i=0;i<=n;i++)for(j=0;i+j<=n;j++){if(i==0&&j==0)dp[i][j]=0;else if(i==0&&j!=0)dp[i][j]=max(dp[i][j],dp[i][j-1]+v[n-j+1]*(i+j));else if(i!=0&&j==0)dp[i][j]=max(dp[i][j],dp[i-1][j]+v[i]*(i+j));elsedp[i][j]=max(dp[i-1][j]+v[i]*(i+j),dp[i][j-1]+v[n-j+1]*(i+j));}int ans=0;for(i=0;i<=n;i++)if(dp[i][n-i]>ans)ans=dp[i][n-i];printf("%d/n",ans);}return 0;}