Word Break II

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

class Solution {

public:

    vector<string> wordBreak(string s, unordered_set<string> &dict) 

    {

        vector<string> result;

        if(check(s,dict)==false) return result;

        

        int v[s.length()+1];v[0]=0;

        generate(result,s,dict,v,1);

        return result;

    }

    bool check(string s, unordered_set<string> &dict) 

    {

        bool able[s.length()+1];

        able[0]=true;

        for(int i=1;i<=s.length();i++)

        {

            able[i]=false;

            for(int j=0;j<i;j++)

            if(able[j] && dict.find(s.substr(j,i-j))!=dict.end())

            {

                able[i]=true;

                continue;

            }

        }

        

        return able[s.length()];

    }

    void generate(vector<string>& result,const string& s,const unordered_set<string> &dict,int* v,int vdep)

    {

        if(v[vdep-1]==s.length())

        {

            string snew=s.substr(0,v[1]);

            for(int i=2;i<vdep;i++)

            {

                snew=snew+" "+s.substr(v[i-1],v[i]-v[i-1]);

            }

            result.push_back(snew);

            return;

        }

        for(int i=1;i<=s.length()-v[vdep-1];i++)

            if(dict.find(s.substr(v[vdep-1],i))!=dict.end())

            {

                v[vdep]=v[vdep-1]+i;

                generate(result,s,dict,v,vdep+1);

            }

    }

};