POJ 3267 The Cow Lexicon

The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7088		Accepted: 3299

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

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

char str[310];
char wrd[610][30];
int w,l,dp[310];    //dp[i]表示从i到 l 所删除的字符数  

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d",&w,&l)){
        scanf("%s",str);
        for(int i=0;i<w;i++)
            scanf("%s",wrd[i]);
        dp[l]=0;
        for(int i=l-1;i>=0;i--){    //从message尾部开始向前检索
            dp[i]=dp[i+1]+1;    //字典单词和str无法匹配时，删除的字符数（最坏的情况）
            for(int j=0;j<w;j++){   //对字典单词枚举 
                int len=strlen(wrd[j]);
                if(len+i<=l && wrd[j][0]==str[i]){  //单词长度小于等于当前待匹配message长度 且单词头字母与信息第i个字母相同
                    int ps=i;   //str的指针
                    int pw=0;   //wrd[j]单词的指针 
                    while(ps<l){    //单词逐字匹配 
                        if(wrd[j][pw]==str[ps++])
                            pw++;
                        if(pw==len){
                            dp[i]=min(dp[i],dp[ps]+(ps-i)-len); //dp[pm]表示从pm到L删除的字符数
                            break;                              //(pm-i)-pd表示从i到pm删除的字符数.则dp[pm]+(pm-i)-pd表示从i到L删除的字符数  
                        }
                    }
                }
            }
        }
        printf("%d
",dp[0]);
    }
    return 0;
}