# POJ 3267 The Cow Lexicon (DP)

The Cow Lexicon

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10284 Accepted: 4914

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

### 思路

dp[i]代表序列[i,n-1]中最少删除的字母个数。

si 为匹配单词结束后的序列游标，i为匹配前的序列游标，len为单词长度

### AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<vector>
#include<queue>

char data[650][50],str[350];
int w,l;
int dp[350];
int solve()
{
dp[l]=0;
for(int i=l-1; i>=0; i--)   //从后向前遍历
{
dp[i]=dp[i+1]+1;        //取最坏情况
for(int j=0; j<w; j++)  //遍历字典
{
if(str[i]==data[j][0])  //第一位相同
{
int si=i,di=0,len=strlen(data[j]);
for(; si<l&&di<len; si++)   //匹配字典中的单词
if(str[si]==data[j][di])
di++;
if(di==len)     //匹配成功
dp[i]=min(dp[i],dp[si]+si-i-len);   //更新dp
}
}
}
return dp[0];
}
int main()
{
while(~scanf("%d%d%*c",&w,&l))
{
gets(str);
for(int i=0; i<w; i++)
gets(data[i]);
printf("%d\n",solve());
}
return 0;
}