Topcoder SRM 392 DIV1 250

Test instructions: Give two strings a,b,a and B all contain lowercase English letters, and both contain and contain only one character ? , you now want the characters in A, b ? Replace each other with other strings (can be different, can be the same, can be empty), make a=b, and require the final A, B string (a=b) the shortest. Or the output is not possible.

Solution: Let a string as ? Number of the previous string, if A's ? is different from the B corresponding position, it is not possible if B's ? The character after the number is different from a corresponding position, it is not possible. Otherwise there must be solutions, violence to find the best solution can. This problem can be a hash or EXTEND-KMP method to achieve the complexity O (N) level, but the implementation of a little trouble, O (n^2) writing simple and easy.

My Code

#include <algorithm>#include <iostream>#include <sstream>#include <cstdlib>#include <cstring>#include <cstdio>#include <string>#include <vector>#include <cmath>#include <set>#include <map>using namespace STD;typedef Long Longllclasstwostringmasks{ Public:stringShortestcommon (stringS1,stringS2) {intT1 =0, t2 =0; for(inti =0; I < s1.length (); i++)if(S1[i] = =' * ') T1 = i; for(inti =0; I < s2.length (); i++)if(S2[i] = =' * ') t2 = i;if(T1 > T2) swap (S1, S2);stringANS1 ="", ANS3 =""; for(inti =0; I < s1.length (); i++)if(S1[i] = =' * ') ans1 = S1.substr (0, i); for(inti =1; I < s2.length (); i++)if(s2[i-1] ==' * ') Ans3 = S2.substr (i, s2.length ()-i); while(s1[0] !=' * '&& s2[0] !=' * ')        {if(s1[0]! = s2[0])return "Impossible"; S1.erase (0,1), S2.erase (0,1); } while(S1.back ()! =' * '&& s2.back ()! =' * ')        {if(S1.back ()! = S2.back ())return "Impossible";        S1.pop_back (), S2.pop_back (); } s1.erase (0,1);stringAns ="."; for(inti =-1; I < (int) S2.length (); i++) {if(I >=0&& S2[i] = =' * ') Break;stringS3; S3.clear (); for(intj =0; J <= I;            J + +) S3 + = S2[j]; S3 + = S1;BOOLOK =1;intT3 =0, t2 =0; while(T3 < s3.length () && T2 < S2.length () && OK) {if(S2[t2] = =' * ') Break;Else if(S3[T3]! = s2[t2]) OK =0;Elset3++, t2++; }if(S2[T2]! =' * ') OK =0;if(ok && (ans = ="."||            Ans.length () > (ans1 + s3 + ans3). Length ()) {ans = ans1 + s3 + ans3; }        }returnAns }};

Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced.

Topcoder SRM 392 DIV1 250