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

O - Treats for the Cows

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

Sample Output

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;
}