[2016-05-04] [Codeforces] [666a-reberland Linguistics]

• Time: 2016-05-04-16:51:34
• Title Number: [2016-05-04][codeforces][666a-reberland linguistics]
• Title: A word consists of a root with a length of not less than 5 and a number of suffixes of 2 or 3, and two adjacent suffixes can not be the same, given a word, ask the total number of suffixes
• Analysis:
  • A DP-like thought, dp[i] Indicates whether the first position to the end of the string can be divided into a legal suffix,
  • So dp[i] = Dp[i +t]∧ (dp[i + 5]∨ substring (i,t)! = substring (i + t, T)); t = = 2 or 3
  • That is, from the backward sweep, each additional character, it is determined by the length of the 2 and 3 is the suffix and the remaining part of the suffix is legitimate
  • Is the new suffix legal,
    • With this suffix, the remainder can still be divided into 2 or 3 suffixes.
    • This suffix is not the same as the next suffix (as long as there is a different division method can be)
      • This suffix is the same as the next suffix length: direct comparison
      • The length is not the same, obviously set up, but to be in the forward number of 5 remaining can be legally divided, that is, dp[i+ 5] for the real situation.
• Problems encountered:
  • When judging whether the current suffix and the next suffix are the same, if the substring is used to compare, I + 2 > length will be wrong at the beginning.
  • The workaround is not to generate substrings, but to find the location of the next and its same string through the Find function
  • Initialize dp[length] = 0;

  
 

   
  

  
#include<iostream>
   
  

  
#include<string>
   
  

  
#include<set>
   
  

  
#include<cstring>
   
  

  
using namespace std;
   
  

  
set<string> s;
   
  

  
const int maxn = 1E4 + 10 ;
   
  

  
int dp[maxn];
   
  

  
int main(){
   
  

  
 string str;
   
  

  
 cin>>str;
   
  

  
 memset(dp,0,sizeof(dp));
   
  

  
 str = str.substr(5);
   
  

  
 int n = str.length();
   
  

  
 string tmp;
   
  

  
 dp[n] = 1;
   
  

  
  for   ( int   i  =   n  -   1  ;   I  >=    0    ;    -- i  ) {  
   
  

  
 for(int l = 2 ; l <= 3 ; ++l){
   
  

  
 tmp = str.substr(i, l);
   
  
 if((Str.Find(tmp,I+L) !=I+L||DP[I+ 5] ) &&DP[I+L] ){
   
  

  
 s.insert(tmp);
   
  

  
 dp[i] = 1;
   
  

  
 }
   
  

  
 }
   
  

  
 }
   
  

  
 cout<<s.size()<<"\n";
   
  

  
 for(set<string>::iterator its = s.begin(); its != s.end();++its){
   
  

  
 cout<<*its<<"\n";
   
  

  
 }
   
  

  
 return 0;
   
  

  
}
  
 

From for notes (Wiz)

[2016-05-04] [Codeforces] [666a-reberland Linguistics]