POJ 3267 The Cow Lexicon

又见面了，还是原来的配方，还是熟悉的DP。。。。直接秒了。。。

The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 7316		Accepted: 3421

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;

int n,m;
char wen[500];
int dp[500];

struct dict
{
    char word[100];
    char theLastLaw;
    int len;
}D[800];

int main()
{
    scanf("%d%d",&n,&m);
    getchar();
    for(int i=1;i<=m;i++)
    {
        scanf("%c",&wen);
    }
    for(int i=0;i<n;i++)
    {
        scanf("%s",D.word);
        D.len=strlen(D.word);
        D.theLastLaw=D.word[D.len-1];
    }
    memset(dp,0,sizeof(dp));
    for(int i=1;i<=m;i++)
    {
        dp=dp[i-1]+1;
        for(int j=0;j<n;j++)
        {
            if(wen==D[j].theLastLaw&&i>=D[j].len)
            {
           //     cout<<i<<": "<<wen<<" catch with: "<<D[j].word<<endl;
                int flag=D[j].len-1,delet=0,pos=-1;
                for(int k=i;k>0;k--)
                {
                    if(wen[k]==D[j].word[flag])
                        flag--;
                    else
                        delet++;
                    if(flag<0){ pos=k; break;}
                }
             //   cout<<"the flag: "<<flag<<"  catch end with: "<<pos<<"  , delet is: "<<delet<<endl;
                if(flag<0)
                {
                    if(pos==-1) pos=0;
                    dp=min(dp,dp[pos-1]+delet);
                }
            }
        }
    }
    printf("%d ",dp);
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )