kuangbin专题十二 POJ3186 Treats for the Cows （区间dp）

Treats for the Cows

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7949		Accepted: 4217

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.

 

 

 

题目大意：给一个长度为n的序列，每次只能从队首或队尾取一个数，第几次取 * f[i] 就是利润，求最大利润。

看到题目果断贪心，只能局部最优，因为是dp专题，但是丝毫不会dp，看了题解发现是区间dp，然后看着理解了一下，

dp[i][j] 表示 从第i个数到第j个数的最大利润，由于只能从dp[i+1][j] 和 dp[i][j-1]到达dp[i][j]，所以状态转移方程可以表示为 dp[i][j] = max(dp[i+1][j] + f[i] * (n-j+i), dp[i][j-1] + f[j] * (n-j+i),

这里是由里向外递推的，逆向遍历i。用n-j+i表示第几次取（可以模拟一个简单的看看）

初始化条件要注意一下，dp[i][i] = f[i] * n 只有一个数时，它就是最后取的。

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
using namespace std;
#define mem(a,b) memset((a),(b),sizeof(a))
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define sz(x) (int)x.size()
#define all(x) x.begin(),x.end()
typedef long long ll;
const int inf = 0x3f3f3f3f;
const ll INF =0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-5;
const ll mod = 1e9+7;
//head
const int maxn = 2000 + 5;
int dp[maxn][maxn], f[maxn];//dp[i][j] 表示从i到j的 最大值

int main() {
    int n;
    while(~scanf("%d", &n)) {
        for(int i = 1; i <= n; i++)
            scanf("%d", &f[i]);
        mem(dp, 0);
        for(int i = 1; i <= n; i++)//初始化 每一个都是最后取的
            dp[i][i] = f[i] * n;
        for(int i = n - 1; i >= 1; i--) {//从里往外推
            for(int j = i + 1; j <= n; j++) {// n - j + i 很巧妙
                dp[i][j] = max(dp[i+1][j] + f[i] * (n - j + i), dp[i][j-1] + f[j] * (n - j + i));//转移方程
            }
        }
        printf("%d
", dp[1][n]);
    }
}