POJ3186 Treats for the Cows —— DP

题目链接：http://poj.org/problem?id=3186

 

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6548		Accepted: 3446

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

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.

Source

USACO 2006 February Gold & Silver

 

 

 

一.正向思维：

1.dp[l][r]表示：左边取了l个， 右边取了r个的最大值。

2.枚举左边取了多少个， 再枚举右边取了多少个。

3.对于当前的 dp[l][r]，它可以是在dp[l-1][r]的基础上取了a[l]；也可以是在dp[l][r-1]的基础上取了a[n+1-r]。所以：

dp[l][r] = max(dp[l-1][r]+a[l], dp[l][r-1]+a[n+1-r])

当然，还需要注意边界条件：l-1>=0，r-1>=0

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 2e3+10;

int n;
int a[MAXN], dp[MAXN][MAXN];

int main()
{
    while(scanf("%d", &n)!=EOF)
    {
        for(int i = 1; i<=n; i++)
            scanf("%d", &a[i]);

        memset(dp, 0, sizeof(dp));
        for(int l = 0; l<=n; l++)
        for(int r = 0; l+r<=n; r++)
        {
            if(l!=0) dp[l][r] = max(dp[l-1][r]+(l+r)*a[l], dp[l][r]);
            if(r!=0) dp[l][r] = max(dp[l][r], dp[l][r-1]+(l+r)*a[n+1-r]);
        }

        int ans = -INF;
        for(int l = 0; l<=n; l++)
            ans = max(ans, dp[l][n-l]);
        printf("%d\n", ans);
    }
}

 

 

二.逆向思维：

1.逆向推导， 即把过程逆过来，然后就变成了：从中间开始往外取，这样就变成了连续的一段。

2.dp[i][j]表示：区间[i, j]的最大值。

3.枚举区间长度， 然后再枚举起点（终点就确定了）。对于dp[i][j]，它可以是在dp[i+1][j]的基础上取了a[i]，也可以是在dp[i][j-1]的基础上取了a[j]。两者取其大。

 

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 2e3+10;

int n;
int a[MAXN], dp[MAXN][MAXN];

int main()
{
    while(scanf("%d", &n)!=EOF)
    {
        for(int i = 1; i<=n; i++)
            scanf("%d", &a[i]);

        for(int i = 1; i<=n; i++)
            dp[i][i] = a[i]*n;

        for(int len = 2; len<=n; len++)
        for(int i = 1; i+len-1<=n; i++)
        {
            int j = i+len-1;
            dp[i][j] = max(dp[i+1][j]+(n-len+1)*a[i], dp[i][j-1]+(n-len+1)*a[j]);
        }

        printf("%d\n", dp[1][n]);
    }
}

 

三.记忆化搜索：

1.上面的两种方法都要考虑枚举顺序的问题，有时比较不好处理。那么可以用记忆化搜索。

2. 思维与方法二一样，只是写法不同。

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 2e3+10;

int n;
int a[MAXN], dp[MAXN][MAXN];

int dfs(int l, int r)
{
    if(l==r) return n*a[l];
    if(dp[l][r]!=-1) return dp[l][r];
    int k = n-r+l;

    dp[l][r] = max(k*a[l]+dfs(l+1, r), k*a[r]+dfs(l,r-1));
    return dp[l][r];
}

int main()
{
    while(scanf("%d", &n)!=EOF)
    {
        for(int i = 1; i<=n; i++)
            scanf("%d", &a[i]);

        memset(dp, -1, sizeof(dp));
        printf("%d\n", dfs(1,n));
    }
}