BZOJ3230 相似子串[后缀数组+二分+st表]

BZOJ3230 相似子串

给一个串，查询排名i和j的子串longest common suffix和longest common prefix

思路其实还是蛮好想的，就是码起来有点恶心。可以发现后缀拍完序之后的子串可以根据sa自前而后的顺序标子串排名，和统计不同字串那里差不多，自lcp后的是新子串，开始标记排名。实际操作时在每个后缀上标这一后缀开始第一个新子串排名，可证是单调增的，然后查询就二分查出子串开始的后缀也就是位置啦，然后用st表维护子串（所在后缀的）lcp即可。lcs的话就把串反转一下做个排序同理st表维护RMQ。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int N=+,Log=;
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline void read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=(x<<)+(x<<)+(c^),c=getchar();f?x-=x:;
 }
 char s[][N];
 int po[N],n,q,a,b,h1,r1,h2,r2,l1,l2;
 ll q1,q2;

 int cnt[][N],h[][N],sa[][N],rk[][N],y[][N],m,p;
 inline void suffix_sort(int l){ m=,p=;
     for(register int i=;i<=n;++i)++cnt[l][rk[l][i]=s[l][i]];
     for(register int i=;i<=m;++i)cnt[l][i]+=cnt[l][i-];
     for(register int i=n;i;--i)sa[l][cnt[l][rk[l][i]]--]=i;
     for(register int k=;k<n;k<<=,p=){
         for(register int i=n-k+;i<=n;++i)y[l][++p]=i;
         for(register int i=;i<=n;++i)if(sa[l][i]>k)y[l][++p]=sa[l][i]-k;
         for(register int i=;i<=m;++i)cnt[l][i]=;
         for(register int i=;i<=n;++i)++cnt[l][rk[l][y[l][i]]];
         for(register int i=;i<=m;++i)cnt[l][i]+=cnt[l][i-];
         for(register int i=n;i;--i)sa[l][cnt[l][rk[l][y[l][i]]]--]=y[l][i];
         swap(rk,y);rk[l][sa[l][]]=p=;
         for(register int i=;i<=n;++i)rk[l][sa[l][i]]=y[l][sa[l][i-]]==y[l][sa[l][i]]&&y[l][sa[l][i-]+k]==y[l][sa[l][i]+k]?p:++p;
         if(p==n)break;m=p;
     }p=;
     for(register int i=;i<=n;h[l][rk[l][i]]=p,p?--p:,++i)while(s[l][i+p]==s[l][sa[l][rk[l][i]-]+p]&&++p);
 }

 ll rank[N],tot=;
 inline void mark(){for(register int i=;i<=n;++i)rank[i]=tot,tot+=n-sa[][i]+-h[][i];--tot;}
 struct st_table{
     int lcp[N][Log];
     inline void build(int l){
         for(register int i=;i<=n;++i)lcp[i][]=h[l][i];
         for(register int k=;k<=po[n];++k)for(register int i=;i<=n+-(<<k);++i)
         lcp[i][k]=_min(lcp[i][k-],lcp[i+(<<(k-))][k-]);
     }
     inline int RMQ(int l,int r){return _min(lcp[l][po[r-l+]],lcp[r-(<<po[r-l+])+][po[r-l+]]);}
 }pos,neg;//positive,negative

 int main(){//freopen("tmp.txt","r",stdin);
     read(n),read(q),scanf("%s",s[]+);
     for(register int i=;i<=n;++i)po[i]=(i-)==(<<(po[i-]+))?po[i-]+:po[i-];
     for(register int i=;i<=n;++i)s[][i]=s[][n-i+];
     suffix_sort(),suffix_sort(),mark();
     pos.build(),neg.build();
 //    for(register int i=1;i<=n;++i)cerr<<s[1][i];puts("");
 //    for(register int i=1;i<=n;++i)cerr<<i<<" "<<rk[1][i]<<" "<<sa[1][i]<<endl;
     for(register int i=;i<=q;++i){
         read(q1),read(q2);
         if(q1>tot||q2>tot){printf("-1\n");continue;}
         h1=upper_bound(rank+,rank+n+,q1)-rank-,h2=upper_bound(rank+,rank+n+,q2)-rank-;
         l1=sa[][h1],l2=sa[][h2],r1=l1+h[][h1]+q1-rank[h1],r2=l2+h[][h2]+q2-rank[h2];
         if(h1==h2)a=_min(r1,r2)-l1+;
         else{
             if(h1>h2)h1^=h2^=h1^=h2;
             a=pos.RMQ(h1+,h2);if((a>r1-l1+)||(a>r2-l2+))a=_min(r1-l1+,r2-l2+);
         }//get a.
         h1=rk[][n-r1+],h2=rk[][n-r2+];//cerr<<l1<<" "<<r1<<" "<<l2<<" "<<r2<<" "<<h1<<" "<<h2<<endl;
         if(h1==h2)b=r1-_max(l1,l2)+;
         else{
             if(h1>h2)h1^=h2^=h1^=h2;
             b=neg.RMQ(h1+,h2);if((b>r1-l1+)||(b>r2-l2+))b=_min(r1-l1+,r2-l2+);
         }//get b.
         printf("%lld\n",1ll*a*a+1ll*b*b);
     }
     return ;
 }