我从这里重现我的算法,其中解释了其逻辑:
dp[i, j] = same as before num[i] = how many subsequences that end with i (element, not index this time)
have a certain length
for i = 1 to n do dp[i, 1] = 1
for p = 2 to k do // for each length this time num = {0}
for i = 2 to n do
// note: dp[1, p > 1] = 0
// how many that end with the prevIoUs element
// have length p - 1
num[ array[i - 1] ] += dp[i - 1, p - 1] *1*
// append the current element to all those smaller than it
// that end an increasing subsequence of length p - 1,
// creating an increasing subsequence of length p
for j = 1 to array[i] - 1 do *2*
dp[i, p] += num[j]
您可以使用段树或二进制索引树进行优化*1*
和*2*
使用。这些将用于有效处理num
阵列上的以下操作:
对于这两种数据结构来说,这都是微不足道的问题。
这将具有复杂性O(n*k*log S)
,这S
是数组中值的上限。这可能足够,也可能不够。为此O(n*k*log n)
,您需要在运行上述算法之前规范化数组的值。规范化意味着将所有数组值转换为小于或等于的值n
。所以这:
5235 223 1000 40 40
成为:
4 2 3 1 1
这可以通过排序(保留原始索引)来完成。