题意

Bessie the cow has arranged a string of N rocks, each containing a single letter of the alphabet, that she wants to build into a fashionable necklace.

Being protective of her belongings, Bessie does not want to share her necklace with the other cow currently living on her side of the barn. The other cow has a name that is a string of M characters, and Bessie wants to be sure that this length-M string does not occur as a contiguous substring anywhere within the string representing her necklace (otherwise, the other cow might mistakenly think the necklace is for her). Bessie decides to remove some of the rocks in her necklace so that the other cow's name does not appear as a substring. Please help Bessie determine the minimum number of rocks she must remove.

贝西收集了N颗石头，每颗石头上都有一个字母，贝西想把这些石头做成项链。

贝西的身边有另一只奶牛，这只奶牛的名字是一个长度为M的字符串，贝西不希望这只牛的名字出现在她的项链上(项链的子串)，她想知道，最少删掉几颗石头就可以避免这种情况发生。

分析

参照The_Virtuoso的题解。

首先如果用AC自动机做这道题显然要把B串建在AC自动机上(AC自动机上就一个串好像有点浪费qwq)。要想B串不出现在A串中，只要把A串在AC自动机上跑，使它一直不遍历到B串的终止节点就能保证B串不是A串的子串。想要最优解自然要dp，那么就可以定义f[i][j]表示A串的第i个字符匹配到了AC自动机上第j个节点保留的最长长度。对于A串上的每一个字符可以删除或者在AC自动机上往下走，最后用A串总长len减掉max{f[len][i]}就是最小删除数了。

时间复杂度\(O(N M)\)

代码

#include<bits/stdc++.h>
#define rg register
#define il inline
#define co const
template<class T>il T read()
{
    rg T data=0;
    rg int w=1;
    rg char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch=='-')
            w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
    {
        data=data*10+ch-'0';
        ch=getchar();
    }
    return data*w;
}
template<class T>il T read(rg T&x)
{
    return x=read<T>();
}
typedef long long ll;
using std::cerr;
using std::endl;

co int N=1e4+1,M=1e3+1;
char s[N],t[M];
namespace AC
{
    int tot;
    int ch[M][26],val[M];
    int fail[M];
    
    void ins(char s[],int n)
    {
        int u=0;
        for(int i=0;i<n;++i)
        {
            int k=s[i]-'a';
            if(!ch[u][k])
                ch[u][k]=++tot;
            u=ch[u][k];
        }
        val[u]=1;
    }
    
    void getfail()
    {
        std::queue<int>Q;
        for(int i=0;i<26;++i)
            if(ch[0][i])
                Q.push(ch[0][i]);
        while(Q.size())
        {
            int u=Q.front();Q.pop();
            for(int i=0;i<26;++i)
            {
                if(ch[u][i])
                {
                    fail[ch[u][i]]=ch[fail[u]][i];
                    Q.push(ch[u][i]);
                }
                else
                    ch[u][i]=ch[fail[u]][i];
            }
        }
    }
    
    int f[N][M];
    
    void solve(char s[],int n)
    {
        for(int i=0;i<n;++i)
        {
            int k=s[i]-'a';
            for(int j=0;j<=tot;++j)
            {
                if(!val[ch[j][k]])
                    f[i+1][ch[j][k]]=std::max(f[i+1][ch[j][k]],f[i][j]+1);
                if(!val[j])
                    f[i+1][j]=std::max(f[i+1][j],f[i][j]);
            }
        }
        int ans=0;
        for(int i=0;i<=tot;++i)
            ans=std::max(ans,f[n][i]);
        printf("%d\n",n-ans);
    }
}

int main()
{
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    scanf("%s",s);
    scanf("%s",t);
    AC::ins(t,strlen(t));
    AC::getfail();
    AC::solve(s,strlen(s));
    return 0;
}