[LeetCode]Scramble String

class Solution {
//O(n^4) DP with some cut off can pass the large judge
//O(n^2) recursion with some cut off can also pass the large judge 
public:
    bool isScramble(string s1, string s2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s1.size() != s2.size()) return false;
        int n = s1.size();
        if(n == 0) return true;
        vector<vector<vector<bool> > > f(n, vector<vector<bool> >(n, vector<bool>(n+1, false)));
        //initialize
        for(int i = 0; i < n; ++i)
        {
            for(int j = 0; j < n; ++j)
            {
                if(s1[i] == s2[j]) f[i][j][1] = true;
            }
        }
        //dp
        for(int len = 2; len <= n; ++len)
        {
            for(int i = 0; i < n; ++i)
            {
                if(i+len-1 >= n) break;
                for(int j = 0; j < n; ++j)
                {
                    if(j+len-1 >= n) break;
                    for(int k = 1; k < len; ++k)
                    {
                        if(i+k < n && j+k < n) f[i][j][len] = f[i][j][len] || (f[i][j][k] && f[i+k][j+k][len-k]);
                        if(j+len-k < n && i+k < n) f[i][j][len] = f[i][j][len] || (f[i][j+len-k][k] && f[i+k][j][len-k]);
                        if(f[i][j][len] == true) break;
                    }
                }
            }
        }
        return f[0][0][n];
    }
};