BZOJ4892 Tjoi2017dna（后缀数组）

　　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<''||c>'')) c=getchar();return c;}

int gcd(int n,int m){return m==?n:gcd(m,n%m);}

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

int T,n,m,sa[N],sa2[N],rk[N<<],tmp[N<<],lg2[N],cnt[N],h[N],f[N][];

char s[N],s2[N];

void make(int n)

{

    memset(cnt,,sizeof(cnt));

    int m=;

    for (int i=;i<=n;i++) cnt[rk[i]=s[i]]++,m=max(m,(int)s[i]);

    for (int i=;i<=m;i++) cnt[i]+=cnt[i-];

    for (int i=n;i>=;i--) sa[cnt[rk[i]]--]=i;

    for (int k=;k<=n;k<<=)

    {

        int p=;

        for (int i=n-k+;i<=n;i++) sa2[++p]=i;

        for (int i=;i<=n;i++) if (sa[i]>k) sa2[++p]=sa[i]-k;

        memset(cnt,,m+<<);

        for (int i=;i<=n;i++) cnt[rk[i]]++;

        for (int i=;i<=m;i++) cnt[i]+=cnt[i-];

        for (int i=n;i>=;i--) sa[cnt[rk[sa2[i]]]--]=sa2[i];

        memcpy(tmp,rk,sizeof(rk));

        p=;rk[sa[]]=;

        for (int i=;i<=n;i++)

        {

            if (tmp[sa[i]]!=tmp[sa[i-]]||tmp[sa[i]+k]!=tmp[sa[i-]+k]) p++;

            rk[sa[i]]=p;

        }

        if (p==n) break;

        m=p;

    }

    for (int i=;i<=n;i++)

    {

        h[i]=max(h[i-]-,);

        while (s[i+h[i]]==s[sa[rk[i]-]+h[i]]) h[i]++;

    }

    for (int i=;i<=n;i++) f[i][]=h[sa[i]];

    for (int j=;j<;j++)

        for (int i=;i<=n;i++)

        f[i][j]=min(f[i][j-],f[min(i+(<<j-),n)][j-]);

    lg2[]=;

    for (int i=;i<=n;i++)

    {

        lg2[i]=lg2[i-];

        if ((<<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+]],f[y-(<<lg2[y-x+])+][lg2[y-x+]]);

}

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+);n=strlen(s+);

        scanf("%s",s2+);m=strlen(s2+);

        for (int i=;i<=m;i++) s[n+i]=s2[i];

        make(n+m);

        int ans=;

        for (int i=;i<=n-m+;i++)

        {

            int x=i;

            for (int j=;j<=&&x-i+<=m;j++)

            {

                x+=query(x,n+x-i+);

                x++;

            }

            if (x-i+<=m) x+=query(x,n+x-i+);

            if (x-i+>m) ans++;

        }

        cout<<ans<<endl;

    }

    return ;

}