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
Sponsor
dp[i]=min(dp[i],dp[pm]+(pm-i)-len);
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int w,l;
char ch[610];
string vec[610];
int f[610];
int main()
{
cin>>w>>l;
for(int i=0;i<l;i++) cin>>ch[i];
for(int i=0;i<w;i++) cin>>vec[i];
f[l]=0;
for(int i=l-1;i>=0;i--)//从后向前更新
{
f[i]=f[i+1]+1;//当前字符需要删除
for(int j=0;j<w;j++)//遍历每个词
{
int len=vec[j].length();//长度
if(len<=l-i+1&&vec[j][0]==ch[i])
//当前遍历到的字符串中的字符到字符串尾的长度大于等于查询单词的长度
//当前遍历到的字符串中的字符等于该单词的首字母
{
int pm=i;//ch的指针
int pz=0;//vec的指针
while(pm<l)//检索从此向后是否逐字匹配
{
if(vec[j][pz]==ch[pm++])
{
pz++;
}
if(pz==len)
{
f[i]=min(f[i],f[pm]+(pm-i)-len);
//所完成从i向后到l与该单词逐字匹配所以更新维护f[]
break;
}
}
}
}
}
cout<<f[0];
}