poj 3267 The Cow Lexicon

The Cow Lexicon

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9590 Accepted: 4603

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

提示

题意：

很少有人知道，奶牛会有自己的字典和W个（1≤W≤600）词，每一个词由字符“a”到"z"组成。他们的cowmunication交流系统，是基于哞哞叫的基础上，并不能听懂在说什么；有时人们会听到似乎没有任何意义的话。例如，Bessie曾收到一个消息：“browndcodw”。之后才知道想要传达的信息是“browncow”，两个字母“d”是因为受到谷仓的干扰。
奶牛要你帮他们解密接收到的消息（字符“a”到"z"）长度为L（2≤L≤300）。该消息有一些额外的字母，他们希望你用最少删除几个字母，使得一个或多个单词能从字典中提取出来解密（和主串完全匹配）。

思路：

DP，要解析戳我。（我把变量放进main里会有莫名的BUG，答案一直不对，连样例都不能过，放到全局变量瞬间AC，我又学会了一招）

示例程序

Source CodeProblem: 3267Code Length: 839BMemory: 408KTime: 204MSLanguage: GCCResult: Accepted#include <stdio.h>#include <string.h>int n,m,i,len[600],i1,l,k,dp[300];char word[600][30],di[301];int main(){    scanf("%d %d",&n,&m);    scanf("%s",di);    for(i=0;n>i;i++)    {        scanf("%s",word[i]);        len[i]=strlen(word[i]);    }    dp[0]=1;    for(i=1;m>i;i++)    {        dp[i]=dp[i-1]+1;        for(i1=0;n>i1;i1++)        {            if(len[i1]-1>i)            {                continue;            }            l=len[i1]-1;            k=i;            while(k>=0&&l>=0&&k>=l)            {                if(word[i1][l]==di[k])                {                    l--;                }                k--;            }            if(dp[i]>dp[k]+i-k-len[i1]&&l<0)            {                dp[i]=dp[k]+i-k-len[i1];            }        }    }    printf("%d",dp[m-1]);    return 0;}