题目大意:给出一个有N个数字的环状序列,让你求一个长度不超过k的和最大的连续子序列。
环状的处理方法一般都是在序列的后面补多一段序列。用s[i]表示到第i位的序列和,对于终点位置为i的序列而言,其长度不超过k的和最大的连续子序列的起点位置就是满足min{s[j]}(j = i-k+1..i-1)的j。显然很容易构造数据卡朴素O(nk)的算法,求区间[i-k+1,i-1]内最小的s[j]就是RMQ问题,用二叉堆、Sparse Table、线段树之类的算法或数据结构可以做到在O(logk)的时间内完成查询,因此总的时间复杂度可以优化到O(nlogk),对于这题的数据规模而言时限是可以接受的。
但通过单调队列可以将时间复杂度优化到O(n)。对于终点为i的序列,每次插入队列的是s[i-1],显然队尾数值大于等于s[i-1]的元素是可以删除的,这样即可保证队列的单调性。而队首的要求则是位置大于等于i-k。这样对于每个终点为i的数列,队首即为最优起点。而每个位置的元素最多只会进出队列一次,因此可以保证时间复杂度是O(n)。
3144367 2010-11-02 00:19:10 Accepted 3415 609MS 1840K 944B C++ IwfWcf
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