如何使用二进制索引树(BIT)找到一定长度的增加子序列的总数?
实际上这是来自Spoj Online Judge
的问题 示例
假设我有一个数组1,2,2,10
长度为3的增加的子序列是1,2,4
和1,3,4
所以答案是2
。
答案 0 :(得分:10)
让:
dp[i, j] = number of increasing subsequences of length j that end at i
一个简单的解决方案是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]
答案是dp[1, k] + dp[2, k] + ... + dp[n, k]
。
现在,这可行,但对于给定的约束效率不高,因为n
可以达到10000
。 k
足够小,因此我们应该尝试找到摆脱n
的方法。
让我们尝试另一种方法。我们还有S
- 数组中值的上限。让我们尝试找一个与此相关的算法。
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]
这具有复杂性O(n * k * S)
,但我们可以非常轻松地将其缩减为O(n * k * log S)
。我们所需要的只是一个数据结构,它可以让我们有效地汇总和更新范围内的元素:segment trees,binary indexed trees等。