北航计算机夏令营机试（？）简单字符串匹配

题意：

• 连 dfs 都不用的匹配，请参见：https://www.noobdream.com/DreamJudge/Issue/page/1378/

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9+7;
 
int n; string o[1010], s[1010]; string pattern;
 
bool match(int index){
    int i=0, j=0;
    while(i<s[index].length()){
        if(pattern[j]=='['){
            int k=j+1; while(pattern[k]!=']') ++k;
            bool ok=false;
            for(int l=j+1; l<k; ++l)
                if(pattern[l]==s[index][i]){
                    ok=true; break;
                }
            if(!ok) return false;
            j=k+1; ++i;
        }
        else if(pattern[j]!='[' && pattern[j]!=']'){
            if(s[index][i]!=pattern[j]){
//                cout<<"index, i, j = "<<index<<", "<<i<<", "<<j<<endl;
//                cout<<"s[index][i], pattern[j] = "<<s[index][i]<<", "<<pattern[j]<<endl;
                return false;
            }
            ++j; ++i;
        }
    }
    return i==s[index].length() && j == pattern.length();
}
 
int main(){
    ios::sync_with_stdio(false);
    int n; cin>>n;
    for(int i=0; i<n; ++i){
        cin>>o[i]; s[i]=o[i];
        for(int j=0; j<s[i].length(); ++j)
            if(s[i][j]<='Z' && s[i][j]>='A') s[i][j]=char(s[i][j]-'A'+'a');
    }
    cin>>pattern; for(char xx : pattern)
        if(xx<='Z' && xx>='A') xx=char(xx-'A'+'a');
    for(int i=0; i<n; ++i){
        if(match(i))
            cout<<i+1<<" "<<o[i]<<endl;
    }
}