POJ 3267 The Cow Lexicon

解题思路:DP

opt[i]表示到i为止最少需要除去多少字符

  
    

   
     
   
     opt[i]
   
     =
   
     min{opt[i
   
     -
   
     1
   
     ]
   
     +
   
     1
   
     ,opt[j]
   
     +
   
     i
   
     -
   
     j
   
     -
   
     len[k]};
  
    
  
    

   
     (
   
     if
   
      msg[i]
   
     ==
   
     word[k][len[k
   
     ]] && msg[j+1,i]包含word[k]);
  
    

 

NULL

   
     

    
      
    
      #include 
    
      <
    
      iostream
    
      >
    
      

    
      using
    
       
    
      namespace
    
       std;


    
      int
    
       main()
{
 
    
      int
    
       i,j,k,t,w,r,l,len[
    
      600
    
      ],opt[
    
      301
    
      ]
    
      =
    
      {
    
      0
    
      };
 
    
      char
    
       msg[
    
      301
    
      ],word[
    
      600
    
      ][
    
      26
    
      ];
 scanf(
    
      "
    
      %d %d
    
      "
    
      , 
    
      &
    
      w, 
    
      &
    
      l);
 scanf(
    
      "
    
      %s
    
      "
    
      , msg);
 
    
      for
    
      (i
    
      =
    
      0
    
      ;i
    
      <
    
      w;i
    
      ++
    
      ){scanf(
    
      "
    
      %s
    
      "
    
      , word[i]);len[i]
    
      =
    
      strlen(word[i]);}
 
    
      for
    
      (i
    
      =
    
      1
    
      ;i
    
      <=
    
      l;i
    
      ++
    
      )
 {
 opt[i]
    
      =
    
      opt[i
    
      -
    
      1
    
      ]
    
      +
    
      1
    
      ;
 
    
      for
    
       (j
    
      =
    
      0
    
      ;j
    
      <
    
      w;j
    
      ++
    
      )
 {
 
    
      if
    
      (i
    
      <
    
      len[j]
    
      ||
    
      msg[i
    
      -
    
      1
    
      ]
    
      !=
    
      word[j][len[j]
    
      -
    
      1
    
      ])
    
      continue
    
      ;
 
    
      for
    
      (k
    
      =
    
      i
    
      -
    
      1
    
      ,t
    
      =
    
      len[j]
    
      -
    
      1
    
      ;k
    
      >=
    
      0
    
      &&
    
      t
    
      >=
    
      0
    
      &&
    
      (i
    
      -
    
      k
    
      -
    
      len[j]
    
      +
    
      t)
    
      <
    
      opt[i];k
    
      --
    
      )
 
    
      if
    
      (msg[k]
    
      ==
    
      word[j][t])t
    
      --
    
      ;
 r
    
      =
    
      t
    
      <
    
      0
    
      ?
    
      (i
    
      -
    
      1
    
      -
    
      k
    
      -
    
      len[j]):i;
 
    
      if
    
      (opt[k
    
      +
    
      1
    
      ]
    
      +
    
      r
    
      <
    
      opt[i])opt[i]
    
      =
    
      opt[k
    
      +
    
      1
    
      ]
    
      +
    
      r;
 }
 }
 printf(
    
      "
    
      %d\n
    
      "
    
      ,opt[l]);
 
    
      return
    
       
    
      0
    
      ;
}