[BZOJ 1652][USACO 06FEB]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…

裸的区间dp,设f[i][j]为区间(i,j)的答案,转移是f[i][j]=max(f[i+1][j]+a[i](n-j+i),f[i][j-1]+a[j]*(n-j+i)); #include<iostream> #include<cstdio> using namespace std; const int N=2005; int n,a[N],f[N][N]; int main() { scanf("%d",&n); for(int i=1;i<…

dp( L , R ) = max( dp( L + 1 , R ) + V_L * ( n - R + L ) , dp( L , R - 1 ) + V_R * ( n - R + L ) ) 边界 : dp( i , i ) = V[ i ] * n -------------------------------------------------------------------------------------------- #include<cstdio> #include&l…

题目 1652: [Usaco2006 Feb]Treats for the Cows Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 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 se…

题目描述 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 re…

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…

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 da…

Treats for the Cows 先搬中文 Descriptions: 给你n个数字v(1),v(2),...,v(n-1),v(n),每次你可以取出最左端的数字或者取出最右端的数字,一共取n次取完.假设你第i次取的数字是x,你可以获得i*x的价值.你需要规划取数顺序,使获得的总价值之和最大. Input 第一行一个数字n(1<=n<=2000). 下面n行每行一个数字v(i).(1<=v(i)<=1000) Output 输出一个数字,表示最大总价值和. Sample In…

题目传送门 做完A Game以后找道区间dp练练手...结果这题没写出来(哭). 和A Game一样的性质,从两边取,但是竟然还有天数,鉴于之前做dp经常在状态中少保存一些东西,所以这次精心设计了状态(不对的). 开始的naive想法:设f[i][j][0/1]为把第1~i / i~n的零食兜售完所得的最多钱,然后我们就会发现这个状态十分难转移. 水题还看了题解. 十分朴素的区间dp,和我在能量项链总结的如出一辙. 本题突破口:可以直接通过区间来得出当前是第几天!! Code #include<…

我是个傻逼,这么水的题都会T #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define R(a,b,c) for(register int a = (b); a <= (c); ++ a) #define nR(a,b,c) for(register int a = (b); a >= (…