The Cow Lexicon
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 8325 | Accepted: 3934 |
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
给你一个模板串，然后给你此传的长度，给你n个单词，让你求最少在模板串上删除多少个字母使得，此模板串刚好能够由给出的单词组成。
记录下每个单词最后一个元素的位置和长度，从模板串开始位置往后找，如果模板串某个位置的元素等于某个单词最后一个字符，那么从这个单词开始往前匹配，如果匹配成功，进行判断是否更优。
状态转移方程dp[i]=min(dp[i],dp[x-1]+differ);
dp[i]表示在模板串第i个位置需要删除最少的元素个数，x-1代表模板串向前x个位置才匹配单词成功，differ表示在匹配的过程中与单词不匹配的个数。
//160K 0MS
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[1007];
char s[307];
struct S
{
char word[27];//单词
int len;//单词长度
char lastword;//单词最后一个元素
}p[607];
int main()
{
int n,len;
while(scanf("%d%d",&n,&len)!=EOF)
{
scanf("%s",s+1);
for(int i=0;i<n;i++)
{
scanf("%s",p[i].word);
p[i].len=strlen(p[i].word);
p[i].lastword=p[i].word[p[i].len-1];
}
memset(dp,0,sizeof(dp));
for(int i=1;i<=len;i++)
{
dp[i]=dp[i-1]+1;//每向后移动就+1
for(int j=0;j<n;j++)
if(p[j].lastword==s[i]&&i>=p[j].len)//如果模板串第i个位置元素等于第j个单词最后的元素且i比此单词长度要长
{
int differ=0,x=i,flag=0;
for(int k=p[j].len-1;x>=1;x--)
if(p[j].word[k]==s[x])
{
k--;
if(k<0){flag=1;break;}//k<0说明此单词全部匹配完
}
else differ++;//在匹配过程中，不相等 的元素个数
if(flag)dp[i]=min(dp[i],dp[x-1]+differ);
}
}
printf("%d\n",dp[len]);
}
return 0;
}