problem

solution

codes

/思路就是对于每个状态下的字符串，枚举可以替换的部分替换作为下一个新的状态。
#include<iostream>
#include<queue>
#include<string>
#include<map>
using namespace std;
int n = 1, flag;
string a, b, ai[1010], bi[1010];
queue<string>q;
map<string, int>ma;//map判重防MLE
int main(){
    cin>>a>>b;
    while(cin>>ai[n]>>bi[n])n++;
    q.push(a);
    ma[a] = 0;
    while(q.size()){
        string t = q.front();  q.pop();
        if(t == b){ flag = 1; break; }
        if(ma[t]>10)break;
        //如果没有这层循环的话，就只能找到第一个子串，后面的会被忽略,如abaaaba abcdaba
        for(int j = 0; j < t.size(); j++){ 
            string nt = t.substr(j);
            for(int i = 1; i < n; i++){
                int tt = nt.find(ai[i]);
                if(tt == string::npos)continue;
                tt += j; //边界条件调起来很麻烦，以及最后直接+j就好了
                string ttt = t.substr(0,tt)+bi[i]+t.substr(tt+ai[i].size());
                if(!ma.count(ttt)){
                    ma[ttt] = ma[t]+1;
                    q.push(ttt);
                }
            }
        }
    }
    if(flag)cout<<ma[b]<<"\n";
    else cout<<"NO ANSWER!\n";
    return 0;
}

#include<iostream>
#include<string>
#include<queue>
#include<set> //set判重
#define maxn 1010
using namespace std;
string ai[maxn],bi[maxn];
struct node{
    string str;
    int st;
    node(string a, int b):str(a),st(b){}
};
set<string>s;
int ans;
int main(){
    string a, b;
    cin>>a>>b;
    int n = 1; 
    while(cin>>ai[n]>>bi[n])n++;
    queue<node>q;
    q.push(node(a,0));
    while(q.size()){
        node t = q.front(); q.pop();
        if(t.str == b){ ans = t.st; break;}
        if(t.st > 10){ ans = 20; break; }
        for(int j = 0; j < t.str.size(); j++){
            string nt = t.str.substr(j);
            for(int i = 1; i < n; i++){
                int tt = nt.find(ai[i]);
                if(tt==string::npos)continue;
                tt += j;
                string ttt = t.str.substr(0,tt)+bi[i]+t.str.substr(tt+ai[i].size());
                if(!s.count(ttt)){
                    q.push(node(ttt,t.st+1));
                    s.insert(ttt);
                }
            }
        }
    }
    //如果<10就因为找不到出来也是不成立的, 即ans没有被赋过值的话
    if(ans == 20 || ans == 0)cout<<"NO ANSWER!\n";
    else cout<<ans<<"\n";
    return 0;
}