[BZOJ2534]Uva10829L-gap字符串(后缀数组+st表)
2017-04-18 08:59
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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 16 int L,n,m; char s1 ,s2 ; int *x,*y,X ,Y ,c ,sa1 ,sa2 ,height1 ,height2 ,rank1 ,rank2 ; int st1 [sz+3],st2 [sz+3],lg ; 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); }
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