思路:首先可以写出n^3dp的状态转移方程:f[i][j]=max{f[j][k]+val[i]},f[i][j]表示最后一步跳到点从j点跳到i点的最大价值(状态不能设成f[i],因为j对后面的决策是有影响的),然后枚举k转移,但这样在时限内是无法通过的,于是考虑如何优化dp,可以改变一下枚举顺序,也就是一般的都是先枚举i再枚举j,可以先枚举j再枚举i,这样有什么好处呢,那么k就以直接用一个指针从j-1扫到1,因为随着i的不断增加,i与j之间的距离是递增的,那么之前合法的决策现在也一定合法,那么…

3315: [Usaco2013 Nov]Pogo-Cow Time Limit: 3 Sec  Memory Limit: 128 MBSubmit: 143  Solved: 79[Submit][Status] Description In an ill-conceived attempt to enhance the mobility of his prize cow Bessie, Farmer John has attached a pogo stick to each of Bes…

3315: [Usaco2013 Nov]Pogo-Cow Time Limit: 3 Sec  Memory Limit: 128 MBSubmit: 352  Solved: 181[Submit][Status][Discuss] Description In an ill-conceived attempt to enhance the mobility of his prize cow Bessie, Farmer John has attached a pogo stick to e…

bzoj1742[Usaco2005 nov]Grazing on the Run 边跑边吃草 bzoj3074[Usaco2013 Mar]The Cow Run 题意: 数轴上有n棵草,牛初始在L位置(bzoj3074的牛初始在1位置),每移动一个单位需要+1s.而每过1s没吃的草腐败度会+1,问吃完所有草的最小腐败度.n≤1000. 题解: 很神的dp.f[l][r][0/1]表示从第l棵草吃到第r棵草,之后到达l/r.则 f[l][r][0]=min(dfs(l+1,r,0)+(n-r+…

3314: [Usaco2013 Nov]Crowded Cows Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 86  Solved: 61[Submit][Status] Description Farmer John's N cows (1 <= N <= 50,000) are grazing along a one-dimensional fence. Cow i is standing at location x(i) and has…

我真想吐槽USACO的数据弱..= = O(n^3)都能A....上面一个是O(n²), 一个是O(n^3) O(n^3)做法, 先排序, dp(i, j) = max{ dp(j, p) } + w( i ) ( t <= p <= j ) 表示跳到第 i 个点, 上一个点是在 j 的最大得分, 其中t是满足条件的最小p. 我们在计算dp(i, j) (1 <= j <= i )时会发现, 随着 j 的递减, t也在不断地减小, 这样我们只要在dp过程中维护h(i, j)表示 m…

3314: [Usaco2013 Nov]Crowded Cows Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 111  Solved: 79[Submit][Status][Discuss] Description Farmer John's N cows (1 <= N <= 50,000) are grazing along a one-dimensional fence. Cow i is standing at location x(…

1640: [Usaco2007 Nov]Best Cow Line 队列变换 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 493  Solved: 253[Submit][Status] Description FJ打算带着他可爱的N (1 ≤ N ≤ 2,000)头奶牛去参加”年度最佳老农”的比赛.在比赛中,每个农夫把他的奶牛排成一列,然后准备经过评委检验. 比赛中简单地将奶牛的名字缩写为其头字母(the initial letter of e…

从左到右扫一遍, 维护一个单调不递减队列. 然后再从右往左重复一遍然后就可以统计答案了. ---------------------------------------------------------------------------- #include<bits/stdc++.h>   #define rep(i, n) for(int i = 0; i < n; ++i) #define clr(x, c) memset(x, c, sizeof(x)) #define for…

1640: [Usaco2007 Nov]Best Cow Line 队列变换 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 543  Solved: 278[Submit][Status] Description FJ打算带着他可爱的N (1 ≤ N ≤ 2,000)头奶牛去参加”年度最佳老农”的比赛.在比赛中,每个农夫把他的奶牛排成一列,然后准备经过评委检验. 比赛中简单地将奶牛的名字缩写为其头字母(the initial letter of e…