The Cow Lexicon

The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10097		Accepted: 4847

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
简单的dp问题，理解dp算法的思想就很好做了

#include<stdio.h>
#include<string.h>
#define N 1000
char str[N],s[N][N];
int dp[N];
int n,m;
void DP()
{
    int len;
    dp[0] = 1;      //从前往后判断，第一个字符dp值设为1
    for(int i = 1;i < m;i++)
    {
        dp[i] = dp[i-1] + 1;  //每次dp值设为当前字符段所能删除的最大值
        for(int j = 0;j < n;j++)
        {
            int l = strlen(s[j]) - 1;
            len = i;
            while(l >= 0 && len >= 0 && len >= l)    //进行匹配
            {
                   if(s[j][l] == str[len])
                   {
                       l--;
                   }
                   len--;
            }
            if(l < 0)              //如果匹配成功，则更新dp值
            {
                if(dp[i] > dp[len] + i - len - strlen(s[j]))      //dp[len] + i - len - strlen(s[j])当前长度减去已匹配的长度，还有剩余未判断的，加剩余的dp值
                {
                    dp[i] = dp[len] + i - len - strlen(s[j]);
                }
            }
        }
    }
}
int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        scanf("%s",str);
        for(int i = 0;i < n;i++)
        {
            scanf("%s",s[i]);
        }
        DP();
        printf("%d\n",dp[m-1]);
    }
    return 0;
}