[BZOJ2534]Uva10829L-gap字符串（后缀数组+st表）

题目描述

传送门

题目大意：求字符串s中有多少子串，满足形如ABA形式，其中A是非空字符串，且B的长度正好为L

题解

这道题和股市的预测实际上时一样的…不过现在忘得快干净了…
B的长度已知是L，首先枚举A的长度i
然后将整个字符串按照长度为i分块，枚举每一个块的端点，设为l，令r=l+i+m，然后对l和r求lcp和lcs，可以发现长度为i+L+i的子串在长度为lcp+lcs的范围内滑动都是合法的，所以对当前答案的贡献为len-i+1
注意滑块不能滑到别的块去，避免重复计算，并且当lcp和lcs都不为0时l被统计了2遍，应该减去

代码

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<cmath>using namespace std;#define N 50005#define sz 16int L,n,m;char s1[N],s2[N];int *x,*y,X[N],Y[N],c[N],sa1[N],sa2[N],height1[N],height2[N],rank1[N],rank2[N];int st1[N][sz+3],st2[N][sz+3],lg[N];long long ans;void build_sa(char *s,int *sa){    m=200;    x=X,y=Y;    for (int i=0;i<m;++i) c[i]=0;    for (int i=0;i<n;++i) ++c[x[i]=s[i]];    for (int i=1;i<m;++i) c[i]+=c[i-1];    for (int i=n-1;i>=0;--i) sa[--c[x[i]]]=i;    for (int k=1;k<=n;k<<=1)    {        int p=0;        for (int i=n-k;i<n;++i) y[p++]=i;        for (int i=0;i<n;++i) if (sa[i]>=k) y[p++]=sa[i]-k;        for (int i=0;i<m;++i) c[i]=0;        for (int i=0;i<n;++i) ++c[x[y[i]]];        for (int i=1;i<m;++i) c[i]+=c[i-1];        for (int i=n-1;i>=0;--i) sa[--c[x[y[i]]]]=y[i];        swap(x,y);        p=1;x[sa[0]]=0;        for (int i=1;i<n;++i)            x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&(sa[i-1]+k<n?y[sa[i-1]+k]:-1)==(sa[i]+k<n?y[sa[i]+k]:-1)?p-1:p++;        if (p>=n) break;        m=p;    }}void build_height(char *s,int *rank,int *height,int *sa){    for (int i=0;i<n;++i) rank[sa[i]]=i;    int k=0;    for (int i=0;i<n;++i)    {        if (!rank[i]) continue;        int j=sa[rank[i]-1];        if (k) --k;        while (s[i+k]==s[j+k]&&i+k<n&&j+k<n) ++k;        height[rank[i]]=k;    }}void rmq(){    for (int i=1,p=0;i<=n;++i)    {        while ((1<<p)<=i) ++p;        lg[i]=p-1;    }    for (int i=0;i<n;++i) st1[i+1][0]=height1[i],st2[i+1][0]=height2[i];    for (int j=1;j<sz;++j)        for (int i=1;i<=n;++i)            if (i+(1<<j)-1<=n)            {                st1[i][j]=min(st1[i][j-1],st1[i+(1<<(j-1))][j-1]);                st2[i][j]=min(st2[i][j-1],st2[i+(1<<(j-1))][j-1]);            }}int main(){    scanf("%d",&L);    scanf("%s",s1);n=strlen(s1);    for (int i=0;i<n;++i) s2[i]=s1[n-i-1];    build_sa(s1,sa1);build_sa(s2,sa2);    build_height(s1,rank1,height1,sa1);    build_height(s2,rank2,height2,sa2);    rmq();    for (int j=1;j<=(n-L)/2;++j)        for (int i=1;i<=n&&i+j+L<=n;i+=j)        {            int l=i,r=i+j+L,ll,rr,lcp,lcs,k;            ll=rank1[l-1]+1,rr=rank1[r-1]+1;            if (ll>rr) swap(ll,rr);            ++ll;            k=lg[rr-ll+1];            lcp=min(st1[ll][k],st1[rr-(1<<k)+1][k]);            lcp=min(lcp,j);            ll=rank2[n-l]+1,rr=rank2[n-r]+1;            if (ll>rr) swap(ll,rr);            ++ll;            k=lg[rr-ll+1];            lcs=min(st2[ll][k],st2[rr-(1<<k)+1][k]);            lcs=min(lcs,j);            int len;            if (lcp&&lcs) len=lcp+lcs-1;            else len=lcp+lcs;            if (len>=j) ans+=(long long)(len-j+1);        }    printf("%lld\n",ans);}