BZOJ 1652: [Usaco2006 Feb]Treats for the Cows

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.

约翰经常给产奶量高的奶牛发特殊津贴，于是很快奶牛们拥有了大笔不知该怎么花的钱．为此，约翰购置了N(1≤N≤2000)份美味的零食来卖给奶牛们．每天约翰售出一份零食．当然约翰希望这些零食全部售出后能得到最大的收益．这些零食有以下这些有趣的特性：

•零食按照1．．N编号，它们被排成一列放在一个很长的盒子里．盒子的两端都有开口，约翰每

天可以从盒子的任一端取出最外面的一个．

•与美酒与好吃的奶酪相似，这些零食储存得越久就越好吃．当然，这样约翰就可以把它们卖出更高的价钱．

•每份零食的初始价值不一定相同．约翰进货时，第i份零食的初始价值为Vi(1≤Vi≤1000)．

•第i份零食如果在被买进后的第a天出售，则它的售价是vi×a．

Vi的是从盒子顶端往下的第i份零食的初始价值．约翰告诉了你所有零食的初始价值，并希望你能帮他计算一下，在这些零食全被卖出后，他最多能得到多少钱．

Sample Output

43

OUTPUT DETAILS:

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.

题解

这道题用区间DP,f[i][j]表示最后i-j+1天卖i~j的商品的最大收益。转移见代码。

代码

 1 /*Author:WNJXYK*/
 2 #include<cstdio>
 3 using namespace std;
 4 
 5 #define LL long long
 6 #define Inf 2147483647
 7 #define InfL 10000000000LL
 8 
 9 inline int abs(int x){if (x<0) return -x;return x;}
10 inline int abs(LL x){if (x<0) return -x;return x;}
11 inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
12 inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
13 inline int remin(int a,int b){if (a<b) return a;return b;}
14 inline int remax(int a,int b){if (a>b) return a;return b;}
15 inline LL remin(LL a,LL b){if (a<b) return a;return b;}
16 inline LL remax(LL a,LL b){if (a>b) return a;return b;}
17 inline void read(int &x){x=0;int f=1;char ch=getchar();while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}x=x*f;}
18 inline void read(LL &x){x=0;LL f=1;char ch=getchar();while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}x=x*f;}
19 inline void read(int &x,int &y){read(x);read(y);}
20 inline void read(LL &x,LL &y){read(x);read(y);}
21 inline void read(int &x,int &y,int &z){read(x,y);read(z);}
22 inline void read(int &x,int &y,int &n,int &m){read(x,y);read(n,m);}
23 inline void read(LL &x,LL &y,LL &z){read(x,y);read(z);}
24 inline void read(LL &x,LL &y,LL &n,LL &m){read(x,y);read(n,m);}
25 
26 const int Maxn=2000;
27 int n,v[Maxn+10];
28 int f[Maxn+5][Maxn+5];
29 int main(){
30     read(n);
31     for (int i=1;i<=n;i++) read(v[i]);
32     for (int i=1;i<=n;i++)
33         f[i][i]=v[i]*n;
34     for (int k=1;k<n;k++){
35         for (int i=1;i<=n-k;i++){
36             int j=i+k;
37             f[i][j]=remax(v[i]*(n-k)+f[i+1][j],v[j]*(n-k)+f[i][j-1]);
38         }
39     }
40     printf("%d\n",f[1][n]);
41     return 0;
42 }

View Code

原文地址：http://www.cnblogs.com/WNJXYK/p/4065601.html

BZOJ 1652: [Usaco2006 Feb]Treats for the Cows

1652: [Usaco2006 Feb]Treats for the Cows

Description

Input

Output

Sample Input

Sample Output