POJ 3267 The Cow Lexicon

The Cow Lexicon
Time Limit: 2000 MS Memory Limit: 65536 K
Total Submissions: 6548 Accepted: 3021
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 ed message (also containing only characters in the range 'A '.. 'Z') of length L (2 ≤ L ≤300) characters that is a bit garbled. in particle, 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
I didn't have any idea when I was doing this. I felt that I was using dp but I wouldn't do it. So I took a look at the problem-solving report.
I think this is quite good.
# Include <iostream>
# Include <string. h>
Using namespace std;
Int dp [400];
String s1 [610];
String s2;
Int min1;
Int main ()
{
Int I, j, n, m, s, t;
Int len, l, x, y;
Cin> n> len;
Cin> s2;
For (I = 1; I <= n; I ++)
{
Cin> s1 [I];
}
Dp [len] = 0;
For (I = len-1; I> = 0; I --)
{
Dp [I] = dp [I + 1] + 1;
For (j = 1; j <= n; j ++)
{
L = s1 [j]. size ();
If (l <= (len-I) & s2 [I] = s1 [j] [0])
{
For (x = I, y = 0; x <= len-1; x ++)
{
If (s2 [x] = s1 [j] [y])
{
Y ++;
}
If (y = l)
{
If (dp [I]> (dp [x + 1] + (x + 1)-I-l ))
{
Dp [I] = (dp [x + 1] + (x + 1)-I-l );
}
}
}
}
}
}
Cout <dp [0] <endl;
Return 0;
}