codeforce 748B

/*通过map将其对应的字母对应起来，注意：1.该字母是否已经对应过（此题的关键，博主用了3个map，第二个map统计是否已经对应过） 2.相同字母的处理方式（我选择一起处理因为相同字母为不同的一种特例）*/ #include <iostream> #include <cstdio> #include <cstring> #include <map> #include <algorithm> using namespace std; int main() { string s1, s2; while(cin >> s1 >> s2){ map <char, char>mp; map <char, int>mp1; map <char, int>mp2; int flag = 0; int cnt = 0; for(int i = 0; i < s1.length(); i++){ mp1[s1[i]]++; mp1[s2[i]]++; if(s1[i] == s2[i])mp1[s1[i]]--; if(mp1[s1[i]] != mp1[s2[i]]){ flag = 1; break; } else{ if(mp1[s1[i]] != 1){ //主要是这里如果第二次匹配需要在这里判断然后在对应，不能先对应后判断，那样永远找不到 if(mp[s1[i]] != s2[i] || mp[s2[i]] != s1[i]){ // cout << s1[i] << " " << s2[i] << endl; flag = 1; break; } } mp[s1[i]] = s2[i]; mp[s2[i]] = s1[i]; if(s1[i] != s2[i] && mp1[s1[i]] == 1){ cnt++; } } } if(flag){ cout << "-1" << endl; continue; } cout << cnt << endl; for(int i = 0; i < s1.length(); i++){ mp2[s1[i]]++; mp2[s2[i]]++; if(s1[i] != s2[i] && mp2[s1[i]] == 1 && mp2[s2[i]] == 1){ cout << s1[i] << " " << s2[i] << endl; } } } return 0; }