问题
How can I find the total number of Increasing sub-sequences of certain length with Binary Index Tree(BIT)?
Actually this is a problem from Spoj Online Judge
Example
Suppose I have an array 1,2,2,10
The increasing sub-sequences of length 3 are 1,2,4
and 1,3,4
So, the answer is 2
.
回答1:
Let:
dp[i, j] = number of increasing subsequences of length j that end at i
An easy solution is in O(n^2 * k)
:
for i = 1 to n do
dp[i, 1] = 1
for i = 1 to n do
for j = 1 to i - 1 do
if array[i] > array[j]
for p = 2 to k do
dp[i, p] += dp[j, p - 1]
The answer is dp[1, k] + dp[2, k] + ... + dp[n, k]
.
Now, this works, but it is inefficient for your given constraints, since n
can go up to 10000
. k
is small enough, so we should try to find a way to get rid of an n
.
Let's try another approach. We also have S
- the upper bound on the values in our array. Let's try to find an algorithm in relation to this.
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]
// 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
dp[i, p] += num[j]
This has complexity O(n * k * S)
, but we can reduce it to O(n * k * log S)
quite easily. All we need is a data structure that lets us efficiently sum and update elements in a range: segment trees, binary indexed trees etc.
来源:https://stackoverflow.com/questions/15057591/how-to-find-the-total-number-of-increasing-sub-sequences-of-certain-length-with