对每个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。当然也可以二分+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 "; #else const char LL[]="%lld "; #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; }