思路：

暴力+pruning。。。




A string such as "word" contains the following abbreviations:

["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "w1r1", "1o2", "2r1", "3d", "w3", "4"]
Given a target string and a set of strings in a dictionary, find an abbreviation of this target string with the smallest possible length such that it does not conflict with abbreviations of the strings in the dictionary.

Each number or letter in the abbreviation is considered length = 1. For example, the abbreviation "a32bc" has length = 4.

Examples:

"apple", ["blade"] -> "a4" (because "5" or "4e" conflicts with "blade")

"apple", ["plain", "amber", "blade"] -> "1p3" (other valid answers include "ap3", "a3e", "2p2", "3le", "3l1").
 

Constraints:

In the case of multiple answers as shown in the second example below, you may return any one of them.
Assume length of target string = m, and dictionary size = n. You may assume that m ≤ 21, n ≤ 1000, and log2(n) + m ≤ 20.













class Solution {
public:
    vector<pair<string,int>>ab;
    unordered_map<string,string>hash;
    string minAbbreviation(string target, vector<string>& dictionary) {        
        vector<string>A;
        for(string &w:dictionary){
            if(w.size()==target.size()){
                A.push_back(w);
            }
        }
        if(A.size()==0)return to_string(target.size());
        
        all(target);
        sort(ab.begin(),ab.end(),[](const pair<string,int>&p1,const pair<string,int>&p2){
           return p1.second<p2.second; 
        });
        
        for(pair<string,int>&p:ab){
            string &s=p.first;
            string &hack=hash[s];
            bool good=true;
            
            for(string &w:A){
                bool ok=false;
                for(int i=0;i<hack.size();i++){
                    if(hack[i]!='#'&&hack[i]!=w[i]){
                        ok=true;
                        break;
                    }
                }
                good=good&ok;
                if(!good)break;
            }
            
            
            if(good){
                return s;
            }
        }
        
        
        return to_string(target.size());
    }
    
    
    void all(string& target){//generate all abbreviation for the target string
        int N=target.size();
        for(int state=0;state<(1<<N);state++){
            int cnt=0;string s1="";string s2="";
            int num=0;
            for(int j=0;j<N;j++){
                if((state&(1<<j))==0){//not use char
                    num++;
                    s2.push_back('#');
                }
                else{//use char
                    if(num!=0){
                        s1=s1+(to_string(num));
                        num=0;
                        cnt++;
                    }
                    cnt++;
                    s1.push_back(target[j]);
                    s2.push_back(target[j]);
                }
            }
            if(num!=0){
                s1=s1+(to_string(num));
                cnt++;
            }
            ab.push_back({s1,cnt});
            hash[s1]=s2;
        }
    }
};