USACO 2013 NOV SILVER

USACO 2013 NOV SILVER
1.pogocow

In an ill-conceived attempt to enhance the mobility of his prize cow
Bessie, Farmer John has attached a pogo stick to each of Bessie’s legs.
Bessie can now hop around quickly throughout the farm, but she has not yet
learned how to slow down.

To help train Bessie to hop with greater control, Farmer John sets up a
practice course for her along a straight one-dimensional path across his
farm. At various distinct positions on the path, he places N targets on
which Bessie should try to land (1 <= N <= 1000). Target i is located at
position x(i), and is worth p(i) points if Bessie lands on it. Bessie
starts at the location of any target of her choosing and is allowed to move
in only one direction, hopping from target to target. Each hop must cover
at least as much distance as the previous hop, and must land on a target.

Bessie receives credit for every target she touches (including the initial
target on which she starts). Please compute the maximum number of points
she can obtain.

肯定要先按照横坐标将这些点排序。
其实 O(N^3) 的 DP 是不难想到的：用 f(i, j) 表示从 i 跳到 j 再往后跳的最大得分（因为只考虑单方向所以假设 j > i），则
f(i, j) = w(i)+max{ f(j, k) }, k>j 且 x(k)-x(j) >= x(j)-x(i)
w(i) 表示第 i 个点的权值。
然后优化：
对于上面方程中的 k，我们只要找到第一个满足条件的 k 就会发现 k+1, k+2, …, n 都是满足条件的。
那么我们干脆维护一个区间最大值：用 m(i, j) 表示 max{ f(i, j), f(i, j+1), …, f(i, n) }
于是上面的方程就可以改写成：
f(i, j) = w(i)+m(j, k), k>j 且 k 是满足条件 x(k)-x(j) >= x(j)-x(i) 的第一个点
显然我们可以用二分查找来找到这个 k。
至于 m(i, j) 的维护，因为 k 在 j 的后面所以我们是倒着来 DP（i 和 j 逆序枚举），那么得到一个 f(i, j) 之后就可以更新 m(i, j) 的值了：
m(i, j) = max{ f(i, j), m(i, j+1) }
问题的复杂度降为 O(N2logN)。