The Cow Lexicon

The Cow Lexicon

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 8815 Accepted: 4162

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

求主串与字典之间匹配要删除的字符数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
#define LL long long
using namespace std;

const int INF = 0x3f3f3f3f;

const int MAX = 20001;

int dp[400];//记录从i到L之间要匹配字典要删除的字符数目

char s[400];

char str[650][400];

int len[650];

int main()
{
int n,L;
while(~scanf("%d %d",&n,&L))
{
scanf("%s",s);
for(int i=0;i<n;i++)
{
scanf("%s",str[i]);
len[i]=strlen(str[i]);
}
memset(dp,0,sizeof(dp));
for(int i=L-1;i>=0;i--)
{
dp[i]=dp[i+1]+1;//最坏的情况,将字符删除
for(int j=0;j<n;j++)//将i到L之间的字符与字典进行匹配
{
if(str[j][0]==s[i]&&i+len[j]<=L)
{
int ss=i;
int t=0;
while(ss<L)
{
if(str[j][t]==s[ss++])
{
t++;
}
if(t==len[j])//如果能够匹配,计算最小值
{
dp[i]=min(dp[i],dp[ss]+ss-i-len[j]);//对于ss,ss到L之间的匹配是已经算出来了,dp[ss]+ss-i-len[j]表示i到L之间匹配要删除的字符数,求最小值
break;
}
}
}
}
}
printf("%d\n",dp[0]);
}

return 0;
}