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
dp,很像wikioi黄金宝藏。
#include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> using namespace std; int n,v[2002],f[2002][2002]; void init() { scanf("%d",&n); int i; for(i=1;i<=n;i++) scanf("%d",&v[i]); } void dp() { int i,j,k,t,maxv; for(k=1;k<=n;k++) for(i=1;i+k-1<=n;i++) {t=n-k+1; j=i+k-1; maxv=max(f[i+1][j]+t*v[i],f[i][j-1]+t*v[j]); f[i][j]=max(f[i][j],maxv); } printf("%d\n",f[1][n]); } int main() { init(); dp(); return 0; }