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

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.

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 0x3f3f3f3f

int 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
",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 0x3f3f3f3f

int 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
",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;
	else
		return 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));
				else
					dp[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;
}