POJ 3267 The Cow Lexicon 动态规划

The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6999		Accepted: 3242

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
代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std ;
#define MAXN 610
#define MAX 310
char word[MAX] ;
char a[MAXN][MAXN] ;
int len[MAXN] ,dp[MAXN] ;
int main()
{
	int i , j , n , m ;
	int k , p ;
	while( scanf( "%d%d" ,&n , &m ) != EOF )
	{
		cin >> word ;
		for( i = 1; i <= n ;i++)
		{
			scanf( "%s" , a[i] ) ;
			len[i] = strlen( a[i] ) ;
		}
		memset( dp , 0 ,sizeof(dp) ) ;
		for( i = 1 ; i <= m ; i++)
		{
			dp[i] = dp[i-1] + 1 ; // 如果不匹配就是加一
			for( j = 1 ;j <= n ;j++)
			{
				k = len[j] - 1 ;
				p = i - 1;
				while( k >= 0 && p >= 0)
				{
					if( a[j][k] == word[p] )
					{
						k-- ;
					}
					p-- ;
				}
				if( k < 0 )
				{
					dp[i] = min(dp[i], dp[p + 1] + i - p - 1 - len[j]);   
					// i - p - 1 - len[j] 是第j个单词匹配成功时要去掉的单词数
					// 转移方程 dp[i] = min( dp[i] ,dp[p+1]+i - p - 1 - len[j] )
					// 既 取 （第i 个不匹配） 和 （匹配）两者 去掉字符的最小值
				}
			}
		}
		cout << dp[m] << endl ;
	}
}