Codeforces 510C Fox And Names topological sorting

Source: Internet
Author: User

Codeforces 510C Fox And Names topological sorting

Portal: cf 510D

Given n strings, ask if such an alphabet exists, so that the string can be sorted in the Lexicographic Order. That is, according to the new alphabet, the sorting meets the lexicographic size.


Assuming that the Lexicographic Order is met, we can obtain the relationship between letters based on the existing strings, and then determine whether feasible solutions exist through topological sorting to output arbitrary solutions, therefore, you only need to determine whether a solution exists.

/*************************************** * *************** File Name: a. cpp * Author: kojimai * Create Time: ******************************** * *********************/# include
 
  
# Include
  
   
# Include
   
    
# Include
    
     
# Include
     
      
Using namespace std; # define FFF 105 char s [FFF] [FFF], ans [30]; int in [FFF]; bool link [26] [26]; queue
      
        P; bool solve () {// determines whether feasible solution int cnt = 0 by means of topological sorting; for (int I = 0; I <26; I ++) {if (in [I] = 0) {p. push (I); ans [cnt ++] = 'A' + I ;}} while (! P. empty () {int now = p. front (); p. pop (); for (int I = 0; I <26; I ++) {if (link [now] [I]) {in [I] --; if (in [I] = 0) {p. push (I); ans [cnt ++] = 'A' + I ;}}} ans [26] = '\ 0'; if (cnt <26) return false; elsereturn true;} int main () {int n; cin> n; for (int I = 0; I <n; I ++) cin> s [I]; bool flag = true; memset (link, false, sizeof (link); memset (in, 0, sizeof (in )); for (int I = 0; I <n-1 & flag; I ++) {bool OK = false; int l1 = Strlen (s [I]), l2 = strlen (s [I + 1]); for (int j = 0; j <l1 & j <l2 &&! OK; j ++) {if (s [I] [j]! = S [I + 1] [j]) {// the size of the Lexicographic Order can be compared if the letters in the same position are different. That is, the relative size of the corresponding letter is OK = true; if (! Link [s [I] [j]-'a'] [s [I + 1] [j]-'a']) {in [s [I + 1] [j]-'a'] ++; link [s [I] [j]-'a'] [s [I + 1] [j]-'a'] = true ;}} if (! OK & l1> l2) flag = false; // if the prefix of a string is identical but the length of the first string is large, the two strings must not meet the Lexicographic Order} if (! Flag) {printf ("Impossible \ n");} else {flag = solve (); if (! Flag) printf ("Impossible \ n"); elseprintf ("% s", ans) ;}return 0 ;}
      
     
    
   
  
 


Related Article

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.