Scramble String

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.
Below is one possible representation of s1 = "great":
great
/ \
gr eat
/ \ / \
g r e at
/ \
a t
To scramble the string, we may choose any non-leaf node and swap its two children.
For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat".
rgeat
/ \
rg eat
/ \ / \
r g e at
/ \
a t
We say that "rgeat" is a scrambled string of "great".
Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae".
rgtae
/ \
rg tae
/ \ / \
r g ta e
/ \
t a
We say that "rgtae" is a scrambled string of "great".
Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Solution: 1. 3-dimensional dp. Contributed by yinlinglin. I really appreciate it!
'dp[k][i][j] == true' means string s1(start from i, length k) is a scrambled string of
string s2(start from j, length k).
2. Recursion + pruning.

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.size() != s2.size()) return false;
        int N = s1.size();
        bool dp[N+1][N][N];
        for (int k = 1; k <= N; ++k)
            for (int i = 0; i <= N-k; ++i)
                for (int j = 0; j <= N-k; ++j)
                {
                    dp[k][i][j] = false;
                    if (k == 1) 
                        dp[1][i][j] = (s1[i] == s2[j]);
                    for (int p = 1; p < k && !dp[k][i][j]; ++p)
                        if (dp[p][i][j] && dp[k-p][i+p][j+p] || dp[p][i][j+k-p] && dp[k-p][i+p][j])
                            dp[k][i][j] = true;
                }
        return dp[N][0][0];
    }
};