Codeforces Round #764 E

E. Masha-forgetful

题链
结论就是任何长的串都可以被2，3长度的串表示
后面就是暴力hash和很常规的dp[i]表示前i个是否匹配了

void solve(){
    int n,m;cin>>n>>m;
    tuple<int,int,int>f2[10][10],f3[10][10][10];
    for(int i=1;i<=n;i++){
        string s;cin>>s;
        for(int j=0;j<m;j++){
            if(j+1<m){
                f2[s[j]-'0'][s[j+1]-'0']={j,j+1,i};
            }
            if(j+2<m){
                f3[s[j]-'0'][s[j+1]-'0'][s[j+2]-'0']={j,j+2,i};
            }
        }
    }
    vector<bool>dp(m+1);
    dp[0]=1;
    tuple<int,int,int>k={0,0,0};
    string s;cin>>s;
    for(int i=0;i<m;i++){
        if (!dp[i]) {
            continue;
        }
        if (i + 2 <= m && f2[s[i] - '0'][s[i + 1] - '0'] != k) {
            dp[i + 2] = true;
        }
        if (i + 3 <= m && f3[s[i] - '0'][s[i + 1] - '0'][s[i + 2] - '0'] != k) {
            dp[i + 3] = true;
        }
    }
    if(!dp[m]){
        cout<<-1<<endl;
    }else{
        int i=m;
        vector<tuple<int,int,int>>ans;
        while(i){
            if(i>=2&&dp[i-2]&&f2[s[i-2]-'0'][s[i-1]-'0']!=k){
                ans.push_back(f2[s[i-2]-'0'][s[i-1]-'0']);
                i-=2;
            }else{
                ans.push_back(f3[s[i-3]-'0'][s[i-2]-'0'][s[i-1]-'0']);
                i-=3;
            }
        }
        reverse(ans.begin(), ans.end());
        cout<<ans.size()<<endl;
        for(auto [i,j,k]:ans){
            cout<<i+1<<' '<<j+1<<' '<<k<<endl;
        }
    }
}