题目
1652: [Usaco2006 Feb]Treats for the Cows
Time Limit: 5 Sec Memory Limit: 64 MB
Submit: 234 Solved: 185
[Submit][Status]
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:
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
1
3
1
5
2
Five treats. On the first day FJ can sell either treat #1 (value 1) or
treat #5 (value 2).
Sample Output
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
/*Author:WNJXYK*/
#include<cstdio>
using namespace std; #define LL long long
#define Inf 2147483647
#define InfL 10000000000LL inline int abs(int x){if (x<) return -x;return x;}
inline int abs(LL x){if (x<) return -x;return x;}
inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
inline int remin(int a,int b){if (a<b) return a;return b;}
inline int remax(int a,int b){if (a>b) return a;return b;}
inline LL remin(LL a,LL b){if (a<b) return a;return b;}
inline LL remax(LL a,LL b){if (a>b) return a;return b;}
inline void read(int &x){x=;int f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}x=x*f;}
inline void read(LL &x){x=;LL f=;char ch=getchar();while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}x=x*f;}
inline void read(int &x,int &y){read(x);read(y);}
inline void read(LL &x,LL &y){read(x);read(y);}
inline void read(int &x,int &y,int &z){read(x,y);read(z);}
inline void read(int &x,int &y,int &n,int &m){read(x,y);read(n,m);}
inline void read(LL &x,LL &y,LL &z){read(x,y);read(z);}
inline void read(LL &x,LL &y,LL &n,LL &m){read(x,y);read(n,m);} const int Maxn=;
int n,v[Maxn+];
int f[Maxn+][Maxn+];
int main(){
read(n);
for (int i=;i<=n;i++) read(v[i]);
for (int i=;i<=n;i++)
f[i][i]=v[i]*n;
for (int k=;k<n;k++){
for (int i=;i<=n-k;i++){
int j=i+k;
f[i][j]=remax(v[i]*(n-k)+f[i+][j],v[j]*(n-k)+f[i][j-]);
}
}
printf("%d\n",f[][n]);
return ;
}