The Cow Lexicon（3267）

最新推荐文章于 2018-10-29 22:04:13 发布

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本文介绍了一种通过删除最少字符将乱码信息转换为有效词汇序列的算法。该算法使用动态规划来确定需要删除的最小字符数，使接收到的乱码信息能够组成字典中的词汇序列。

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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

参考:dp的思想

#include <stdio.h> 
#include <string.h>  
#include <math.h>  
#include <stdlib.h>  
#include <ctype.h>
int min(int a,int b)
{
	return a>b?b:a;
}    
int main()  
{    
	int W,L;
	char s[301];
	char dict[601][301];
	int dp[301]={0};
	int i=0,j,k,t,m;
	scanf("%d %d",&W,&L);
	scanf("%s",s);
	for(j=0;j<W;j++)
	{
		scanf("%s",dict[j]);		
	}
	dp[L]=0;
	//从后面开始往前找
	for(i=L-1;i>=0;i--)
	{
		dp[i]=dp[i+1]+1;//后面找不到词典的单词则必须多删一个，若找到则取减的少的那个
		for(j=0;j<W;j++)
		{
			if(strlen(dict[j])<=L-i && dict[j][0]==s[i])
			{
				t=i;
				m=0;
				while(t<L)
				{
					if(dict[j][m]==s[t++])
					m++;
					if(m==strlen(dict[j]))
					{
				 		dp[i]=min(dp[t]+t-i-m,dp[i]);  
                      <span style="white-space:pre">				</span>break;  
					}
				}
			}
		}
	}
	printf("%d\n",dp[0]);  
      
    return 0;  
}