POJ-3267-The Cow Lexicon

The Cow Lexicon

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 10084 Accepted: 4839

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

Source

USACO 2007 February Silver

dp[i]表示从message中第i个字符开始，到第L个字符（结尾处）这段区间所删除的字符数，初始化为dp[L]=0

由于我的程序是从message尾部向头部检索匹配，所以是下面的状态方程：

 

从程序可以看出，第i个位置到L所删除的字符数，总是先取最坏情况，只有可以匹配单词时才进入第二条方程进行状态优化更新。

#include <iostream>#include  <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <algorithm>using namespace std;char zidian[610][30];char str[310];int W,L;int dp[310];int main(){    while(~scanf("%d %d",&W,&L))    {        int i,j;        memset(dp,0,sizeof(dp));        scanf("%s",str);        for(i = 0; i < W; i++)            scanf("%s",zidian[i]);        dp[L] = 0;        for( i = L - 1; i >=0; i--)        {            dp[i] = dp[i+1] + 1;            for( j = 0; j < W; j++ )            {                int len = strlen(zidian[j]);                if(len <= L - 1 && zidian[j][0] == str[i])                {                    int pm = i;                    int pd = 0;                    while(pm < L)                    {                        if(zidian[j][pd] == str[pm++])                            pd++;                        if(pd == len)                        {                            dp[i] = min(dp[i],dp[pm]+(pm - i) - len);                            break;                        }                    }                }            }        }        printf("%d\n",dp[0]);    }    return 0;}