對每個子串暴力匹配至失配三次即可。可以用SA查lcp。然而在bzoj上被卡常了。當然也可以二分+哈希或者SAM甚至FFT。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<‘
 
A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1
 
)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int T,n,m,sa[N],sa2[N],rk[N<<1],tmp[N<<1],lg2[N],cnt[N],h[N],f[N][19];
char s[N],s2[N];
void make(int n)
{
    memset(cnt,0,sizeof(cnt));
    int m=0;
    for (int i=1;i<=n;i++) cnt[rk[i]=s[i]]++,m=max(m,(int)s[i]);
    for (int i=1;i<=m;i++) cnt[i]+=cnt[i-1
 
];
    for (int i=n;i>=1;i--) sa[cnt[rk[i]]--]=i;
    for (int k=1;k<=n;k<<=1)
    {
        int p=0;
        for (int i=n-k+1;i<=n;i++) sa2[++p]=i;
        for (int i=1;i<=n;i++) if (sa[i]>k) sa2[++p]=sa[i]-k;
        memset(cnt,0,m+1<<2);
        for (int i=1;i<=n;i++) cnt[rk[i]]++;
        for (int i=1;i<=m;i++) cnt[i]+=cnt[i-1];
        for (int i=n;i>=1;i--) sa[cnt[rk[sa2[i]]]--]=sa2[i];
        memcpy(tmp,rk,sizeof(rk));
        p=1;rk[sa[1]]=1;
        for (int i=2;i<=n;i++)
        {
            if (tmp[sa[i]]!=tmp[sa[i-1]]||tmp[sa[i]+k]!=tmp[sa[i-1]+k]) p++;
            rk[sa[i]]=p;
        }
        if (p==n) break;
        m=p;
    }
    for (int i=1;i<=n;i++)
    {
        h[i]=max(h[i-1]-1,0);
        while (s[i+h[i]]==s[sa[rk[i]-1]+h[i]]) h[i]++;
    }
    for (int i=1;i<=n;i++) f[i][0]=h[sa[i]];
    for (int j=1;j<19;j++)
        for (int i=1;i<=n;i++)
        f[i][j]=min(f[i][j-1],f[min(i+(1<<j-1),n)][j-1]);
    lg2[1]=0;
    for (int i=2;i<=n;i++)
    {
        lg2[i]=lg2[i-1];
        if ((2<<lg2[i])<=i) lg2[i]++;
    }
}
int query(int x,int y)
{
    x=rk[x],y=rk[y];
    if (x==y) return N;
    if (x>y) swap(x,y);x++;
    return min(f[x][lg2[y-x+1]],f[y-(1<<lg2[y-x+1])+1][lg2[y-x+1]]);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4892.in","r",stdin);
    freopen("bzoj4892.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    T=read();
    while (T--)
    {
        scanf("%s",s+1);n=strlen(s+1);
        scanf("%s",s2+1);m=strlen(s2+1);
        for (int i=1;i<=m;i++) s[n+i]=s2[i];
        make(n+m);
        int ans=0;
        for (int i=1;i<=n-m+1;i++)
        {
            int x=i;
            for (int j=1;j<=3&&x-i+1<=m;j++)
            {
                x+=query(x,n+x-i+1);
                x++;
            }
            if (x-i+1<=m) x+=query(x,n+x-i+1);
            if (x-i+1>m) ans++;
        }
        cout<<ans<<endl;
    }
    return 0;
}

BZOJ4892 Tjoi2017dna（後綴數組）