POJ 3267, The Cow Lexicon

Time Limit: 2000MS  Memory Limit: 65536K
Total Submissions: 2856  Accepted: 1252

Description

Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of the characters 'a'..'z'. Their cowmunication system, based on mooing, is not very accurate; sometimes they hear words that do not make any sense. For instance, Bessie once received a message that said "browndcodw". As it turns out, the intended message was "browncow" and the two letter "d"s were noise from other parts of the barnyard.

The cows want you to help them decipher a received message (also containing only characters in the range 'a'..'z') of length L (2 ≤ L ≤ 300) characters that is a bit garbled. In particular, they know that the message has some extra letters, and they want you to determine the smallest number of letters that must be removed to make the message a sequence of words from the dictionary.

 

Input
Line 1: Two space-separated integers, respectively: W and L
Line 2: L characters (followed by a newline, of course): the received message
Lines 3..W+2: The cows' dictionary, one word per line

 

Output
Line 1: a single integer that is the smallest number of characters that need to be removed to make the message a sequence of dictionary words.

 

Sample Input
6 10
browndcodw
cow
milk
white
black
brown
farmer

 

Sample Output
2

 

Source
USACO 2007 February Silver

---

//  POJ3267.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < string >
using   namespace  std;

int  main( int  argc,  char *  argv[])
{
     char  words[ 600 ][ 26 ];
     char  inw[ 301 ];
     int  W, L;

    scanf( " %d %d\n " , & W, & L);
    gets(inw);
     for  ( int  i  =   0 ; i  <  W;  ++ i)gets(words[i]);

     int  del[ 301 ];
    del[L]  =   0 ;
     for ( int  i  =  L  -   1  ; i  >=   0  ;  -- i)
    {
        del[i]  =   301 ;
         for ( int  j  =   0  ; j  <  W ; j  ++ )
        {
             int  k  =   0 ,l  =  i;
             for (;words[j][k]  &&  l  <  L;  ++ l)
                 if (words[j][k]  ==  inw[l])k ++ ;

             if ( ! words[j][k])
                del[i]  =  std::min(std::min(del[i],del[i  +   1 ]  +   1 ),del[l]  +  l  -  i  -  k);
             else
                del[i]  =  min(del[i], del[i  +   1 ]  +   1 );
        }
    }
    cout << del[ 0 ] << endl;
     return   0 ;
}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/13/1582222.html