The Cow Lexicon

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8211		Accepted: 3864

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.

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

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.

6 10

browndcodw

cow

milk

white

black

brown

farmer

2

USACO 2007 February Silver

给出一个字符串，问最少去掉几个字符得到的序列是全然由给出的字典组成。字典在以下给出

依据问题。能够想到最后得到的字符串是字典组成的，那么组成这个字符串的单词在一開始就分散在给出的串中，我们要找的就是推断单词在字符串的位置。

dp[i]表示从头開始到第i个字符最少要消除多少个字符，

首先dp[0] = 1 ；

dp[i] = dp[i-1] , 匹配不到

dp[i] = dp[j] + k 出现字符串能够匹配到从j+1到i的字符串中，k = 从j+1到i完毕匹配须要消除的字符。

一開始用最长公共子序列写，后来发现，在对dp[i]寻找从0到i有没有单词能够匹配的时候。要求，单词的最后一个单词一定和s[i]同样。不然那个单词在之前就已经匹配过了。

所以仅仅要用普通的遍历就能够了。

PS：对于不知道開始或结束的位置的字符串判字串，用最长公共子序列，假设知道一个确定的位置，直接匹配。

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

struct node

{

    int l ;

    char s[30] ;

} p[700];

int dp[400];

char s[400] ;

int cmp(node a,node b)

{

    return a.l < b.l ;

}

int main()

{

    int i , j , n , l ;

    while( scanf("%d %d", &n, &l)!=EOF)

    {

        scanf("%s", s);

        for(i = 0 ; i < n ; i++)

        {

            scanf("%s", p[i].s);

            p[i].l = strlen(p[i].s);

        }

        sort(p,p+n,cmp);

        memset(dp,0,sizeof(dp));

        dp[0] = 1 ;

        for(i = 0 ; i < l ; i++)

        {

            if(i)

                dp[i] = dp[i-1]+1 ;

            for( j = 0 ; p[j].l <= i+1 ; j++)

            {

                int k1 = i , k2 = p[j].l-1 ;

                if( s[k1] != p[j].s[k2] ) continue ;

                while( k1 >= 0 && k2 >= 0 )

                {

                    if( s[k1] == p[j].s[k2] )

                    {

                        k1-- ;

                        k2-- ;

                    }

                    else

                        k1-- ;

                }

                if( k2 == -1 )

                {

                    dp[i] = min( dp[i],dp[k1]+i-k1-p[j].l );

                }

            }

        }

        printf("%d\n", dp[l-1]);

    }

    return 0;

}