杭电15题 The Cow Lexicon

Problem 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

PKU

题意：给定一个字符串检查最大数量的匹配单词是多少；

思路：

AC代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<algorithm>
 #include<cmath>

 using namespace std;

 int main()
 {
     int dp[]={};
     string str[];
     int w,l;
     cin>>w>>l;//输入w和l
     string s;
     cin>>s;//输入待检测的字符串；
     int i;
     for(i=;i<w;i++)
         cin>>str[i];
     int j;int k;
     dp[l]=;
     int right=;
     int len;
     for(i=l-;i>=;i--)
     {
         dp[i]=dp[i+]+;
         for(j=;j<w;j++)
         {
             if(s[i]==str[j][])
             {//进行匹配；
                 k=i;
                 len=str[j].length();
                 right=;
                 while(right<len&&k<l)
                 {
                     if(s[k]==str[j][right]){right++;}
                     k++;
                     if(right==len)
                     {
                         dp[i]=min(dp[i],dp[k]+k-i-len);
                         break;
                     }
                 }
             }
         }
     }
     cout<<dp[]<<endl;
     return ;
 }