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…

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]…

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…

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  …

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…

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: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…

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…

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[…

题目链接:http://poj.org/problem?id=3186 题目大意:给出的一系列的数字,可以看成一个双向队列,每次只能从队首或者队尾出队,第n个出队就拿这个数乘以n,最后将和加起来,求最大和. 解题思路:有两种写法: ①这是我一开始想的,从外推到内,设立数组dp[i][j]表示剩下i~j时的最优解,则有状态转移方程: dp[i][j]=dp[i][j]=max(dp[i-1][j]+a[i-1]*(n-(j-i+1)),dp[i][j+1]+a[j+1]*(n-(j+1-i)))…