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简单DP,对第i个位置考虑要不要保留。。。。根据以前的状态推最少需要删除几个#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n,m;
char wen[500];
int dp[500];
struct dict
{
char word[100];
char theLastLaw;
int len;
}D[800];
int main()
{
scanf("%d%d",&n,&m);
getchar();
for(int i=1;i<=m;i++)
{
scanf("%c",&wen[i]);
}
for(int i=0;i<n;i++)
{
scanf("%s",D[i].word);
D[i].len=strlen(D[i].word);
D[i].theLastLaw=D[i].word[D[i].len-1];
}
memset(dp,0,sizeof(dp));
for(int i=1;i<=m;i++)
{
dp[i]=dp[i-1]+1;
for(int j=0;j<n;j++)
{
if(wen[i]==D[j].theLastLaw&&i>=D[j].len)
{
// cout<<i<<": "<<wen[i]<<" catch with: "<<D[j].word<<endl;
int flag=D[j].len-1,delet=0,pos=-1;
for(int k=i;k>0;k--)
{
if(wen[k]==D[j].word[flag])
flag--;
else
delet++;
if(flag<0){ pos=k; break;}
}
// cout<<"the flag: "<<flag<<" catch end with: "<<pos<<" , delet is: "<<delet<<endl;
if(flag<0)
{
if(pos==-1) pos=0;
dp[i]=min(dp[i],dp[pos-1]+delet);
}
}
}
}
printf("%d\n",dp[m]);
return 0;
}
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