poj 3267 The Cow Lexicon (动态规划)

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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 10browndcodwcowmilkwhiteblackbrownfarmer

Sample Output

2
dp[i]代表以i为结尾的最优选择,状态转移方程为:

dp[i]=min(dp[i-1]+1,opt); opt:包括第 i个字符的最优选择;

把原先的字典里的词组反转,枚举字典里的全部词汇,找到最优解。

#include<stdio.h>#include<queue>#include<map>#include<string>#include<string.h>using namespace std;#define N 305const int inf=0x1f1f1f1f;char str[N*2][30],s[N];int dp[N];int main(){    int i,j,k,n,m;    char ch;    while(scanf("%d%d",&m,&n)!=-1)    {        scanf("%s",s+1);        for(i=0;i<m;i++)        {            scanf("%s",str[i]);            int len=strlen(str[i]);  //字符串反转            for(j=0;j<len/2;j++)            {                ch=str[i][j];                str[i][j]=str[i][len-j-1];                str[i][len-1-j]=ch;            }            str[i][len]='\0';        }        dp[0]=0;        int tmp;        for(i=1;i<=n;i++)        {            tmp=dp[i-1]+1;  //tmp初始化为该字符舍去时的值            for(j=0;j<m;j++)            {                if(str[j][0]!=s[i])                          continue;                int l=1,len=strlen(str[j]);                for(k=i-1;k>0&&l<len;k--)  //寻找该单词出现的最早位置                {                    if(s[k]==str[j][l])                        l++;                }                if(l==len)      //包括此单词                    tmp=min(tmp,dp[k]+(i-k-len));            }            dp[i]=tmp;        }        printf("%d\n",dp[n]);    }


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