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…

http://poj.org/problem?id=3186 Treats for the Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4041   Accepted: 2063 Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of…

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…

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…

 Treats for the Cows Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3186 Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amounts of…

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…

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…

http://poj.org/problem?id=3186   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…

[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…

P2858 [USACO06FEB]奶牛零食Treats for the Cows区间dp,级像矩阵取数, f[i][i+l]=max(f[i+1][i+l]+a[i]*(m-l),f[i][i+l-1]+a[i+l]*(m-l)); #include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<cmath> #include<ctime&g…

P2858 [USACO06FEB]奶牛零食Treats for the Cows 区间dp 设$f[l][r]$为取区间$[l,r]$的最优解,蓝后倒着推 $f[l][r]=max(f[l+1][r]+a[l]*p,f[l][r-1]+a[r]*p)$ #include<iostream> #include<cstdio> #include<cstring> using namespace std; int max(int a,int b){return a>b…

http://www.lydsy.com/JudgeOnline/problem.php?id=1652 dp.. 我们按间隔的时间分状态k,分别为1-n天 那么每对间隔为k的i和j.而我们假设i或者j在间隔时间内最后取.那么在这个间隔时间内最后取的时间就是n-k+1(这个自己想..也就是说,之前在n-(k-1)+1的时间间隔内取过了,现在我们要多了一个时刻,相当于取这个早了一个时间) 然后就是 k为阶段 i为左端点 j=i+k-1为右端点 t=n-k+1为i-j取最后一个的时间 然后转移 f[…

[USACO06FEB]奶牛零食Treats for the Cows 思路: 区间DP: 代码: #include <bits/stdc++.h> using namespace std; #define maxn 2005 #define ll long long ll n,ai[maxn],dp[maxn][maxn],sum[maxn]; inline void in(ll &now) { ; ')Cget=getchar(); ') { now=now*+Cget-'; Cg…

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…

题目链接  Treats for the Cows 直接区间DP就好了,用记忆化搜索是很方便的. #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define rep(i,a,b) for(int i(a); i <= (b); ++i) #define LL long long + ; LL f[Q]…

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…

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…

题目链接:http://poj.org/problem?id=3186 Treats for the Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6548   Accepted: 3446 Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for giving vast amount…

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…

SP740 TRT - Treats for the Cows 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 tr…

P2858 [USACO06FEB]奶牛零食Treats for the Cows 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 ti…

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…

跟某NOIP的<矩阵取数游戏>很像. f(i,j)表示从左边取i个,从右边取j个的答案. f[x][y]=max(dp(x-1,y)+a[x]*(x+y),dp(x,y-1)+a[n-y+1]*(x+y)). ans=max{f(i,n-i)}. #include<cstdio> #include<algorithm> #include<cstring> using namespace std; #define N 2001 int n,a[N],f[N][…

题目链接:http://poj.org/problem?id=3186 题意:第一个数是N,接下来N个数,每次只能从队列的首或者尾取出元素. ans=每次取出的值*出列的序号.求ans的最大值. 样例 : input:5  1 2 1 5 2 output:43 思路:区间dp,用两个指针i和j代表区间,dp[i][j]表示这个区间的最大值. ///最开始想想着每次拿值最小的就好了,因为之前没接触过区间dp,就这样naive的交了,意料之中的WA了. ///找到一个范例 eg:1000001  …

DP[i][j]表示现在开头是i物品,结尾是j物品的最大值,最后扫一遍dp[1][1]-dp[n][n]就可得到答案了 稍微想一下,就可以, #include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<algorithm> #include<cstring> #include<cstring> #include<vect…

简单DP dp[i][j]表示的是i到j这段区间获得的a[i]*(j-i)+... ...+a[j-1]*(n-1)+a[j]*n最大值 那么[i,j]这个区间的最大值肯定是由[i+1,j]与[i,j-1]区间加上端点的较大值推过来的. #include<cstdio> #include<cstring> #include<cmath> #include<stack> #include<vector> #include<string>…

https://www.luogu.org/problemnew/show/P2858 方程很好想,关键我多枚举了一次(不过也没多大关系) #include <bits/stdc++.h> #define read read() #define up(i,l,r) for(register int i = (l);i <= (r); i++) using namespace std; ; int n,a[N],f[N][N]; int read { ;char ch = getchar(…

蒟蒻许久没做题了,然后连动规方程都写不出了. 参照iwtwiioi大神,这样表示区间貌似更方便. 令f[i, j]表示i到j还没卖出去,则 f[i, j] = max(f[i + 1, j] + v[i] * T, f[i, j - 1] + v[j] * T) (←这样用推的方式更好想一点..) /************************************************************** Problem: User: rausen Language: Pasc…

题目来源:http://poj.org/problem?id=3186 (http://www.fjutacm.com/Problem.jsp?pid=1389) /** 题目意思: 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…