USACO chapter 2 section 2.3 The Longest Prefix

USER: tian tianbing [tbbd4261]
TASK: prefix
LANG: C++
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.000 secs, 6916 KB]
Test 2: TEST OK [0.000 secs, 6916 KB]
Test 3: TEST OK [0.000 secs, 6916 KB]
Test 4: TEST OK [0.000 secs, 6916 KB]
Test 5: TEST OK [0.022 secs, 6916 KB]
Test 6: TEST OK [0.173 secs, 6916 KB]
All tests OK.
Your program ('prefix') produced all correct answers!  This is your
submission #5 for this problem.  Congratulations!



DP:dp[i]=(dp[i]||dp[i-ll]);

   dp[i]           i。      dp[i-ll]    dp[i]     。

  
  
  
  

   
   
   
   /*
   
   
   
   
ID:tbbd4261
PROG:prefix
LANG:C++

   
   
   
   */
   
   
   
   

#include
   
   
   
   <
   
   
   
   iostream
   
   
   
   >
   
   
   
   
#include
   
   
   
   <
   
   
   
   fstream
   
   
   
   >
   
   
   
   
#include
   
   
   
   <
   
   
   
   string
   
   
   
   >
   
   
   
   
#include
   
   
   
   <
   
   
   
   cstring
   
   
   
   >
   
   
   
   

   
   
   
   using
   
   
   
    
   
   
   
   namespace
   
   
   
    std;
ifstream fin(
   
   
   
   "
   
   
   
   prefix.in
   
   
   
   "
   
   
   
   );
ofstream fout(
   
   
   
   "
   
   
   
   prefix.out
   
   
   
   "
   
   
   
   );

   
   
   
   const
   
   
   
    
   
   
   
   int
   
   
   
    MAX
   
   
   
   =
   
   
   
   2000005
   
   
   
   ;

   
   
   
   string
   
   
   
    prefix[
   
   
   
   205
   
   
   
   ];

   
   
   
   char
   
   
   
    s[MAX]
   
   
   
   =
   
   
   
   {
   
   
   
   0
   
   
   
   };

   
   
   
   bool
   
   
   
    dp[MAX]
   
   
   
   =
   
   
   
   {
   
   
   
   0
   
   
   
   };

   
   
   
   int
   
   
   
    cnt
   
   
   
   =
   
   
   
   0
   
   
   
   ,n
   
   
   
   =
   
   
   
   0
   
   
   
   ,i
   
   
   
   =
   
   
   
   0
   
   
   
   ,j
   
   
   
   =
   
   
   
   0
   
   
   
   ,k;


   
   
   
   int
   
   
   
    main()
{ 
    
   
   
   
   while
   
   
   
   (fin
   
   
   
   >>
   
   
   
   prefix[
   
   
   
   ++
   
   
   
   cnt], prefix[cnt]
   
   
   
   !=
   
   
   
   "
   
   
   
   .
   
   
   
   "
   
   
   
   )
                ;
    cnt
   
   
   
   --
   
   
   
   ;
    
   
   
   
   char
   
   
   
    ch; i
   
   
   
   =
   
   
   
   0
   
   
   
   ;
    
   
   
   
   while
   
   
   
   (fin
   
   
   
   >>
   
   
   
   ch)s[
   
   
   
   ++
   
   
   
   i]
   
   
   
   =
   
   
   
   ch;
    n
   
   
   
   =
   
   
   
   i;  
    
   
   
   
   bool
   
   
   
    flag;
    dp[
   
   
   
   0
   
   
   
   ]
   
   
   
   =
   
   
   
   true
   
   
   
   ;
    
   
   
   
   for
   
   
   
   ( i
   
   
   
   =
   
   
   
   1
   
   
   
   ; i
   
   
   
   <=
   
   
   
   n; i
   
   
   
   ++
   
   
   
   )
    {
         
   
   
   
   for
   
   
   
   (j
   
   
   
   =
   
   
   
   1
   
   
   
   ; j
   
   
   
   <=
   
   
   
   cnt; j
   
   
   
   ++
   
   
   
   )
         {
                  flag
   
   
   
   =
   
   
   
   true
   
   
   
   ;
                  
   
   
   
   int
   
   
   
    ll
   
   
   
   =
   
   
   
   prefix[j].length();
                  
   
   
   
   if
   
   
   
   (i
   
   
   
   <
   
   
   
   ll)
   
   
   
   continue
   
   
   
   ;
                  
   
   
   
   for
   
   
   
   (k
   
   
   
   =
   
   
   
   0
   
   
   
   ; k
   
   
   
   <
   
   
   
   ll; k
   
   
   
   ++
   
   
   
   )
                     
   
   
   
   if
   
   
   
   (prefix[j][k]
   
   
   
   !=
   
   
   
   s[i
   
   
   
   -
   
   
   
   ll
   
   
   
   +
   
   
   
   1
   
   
   
   +
   
   
   
   k])
                     {
                             flag
   
   
   
   =
   
   
   
   false
   
   
   
   ; 
   
   
   
   break
   
   
   
   ;
                     }
                     
   
   
   
   if
   
   
   
   (flag)dp[i]
   
   
   
   =
   
   
   
   (dp[i]
   
   
   
   ||
   
   
   
   dp[i
   
   
   
   -
   
   
   
   ll]);
                     if(dp[i])break;
         }  
    }
    
    
   
   
   
   for
   
   
   
   (k
   
   
   
   =
   
   
   
   n; k
   
   
   
   >=
   
   
   
   0
   
   
   
   ; k
   
   
   
   --
   
   
   
   )
             
   
   
   
   if
   
   
   
   (dp[k]) { fout
   
   
   
   <<
   
   
   
   k
   
   
   
   <<
   
   
   
   endl; 
   
   
   
   break
   
   
   
   ;}
    
   
   
   
   return
   
   
   
    
   
   
   
   0
   
   
   
   ;
}