[BZOJ3230]相似子串(后缀数组)

显然可以通过后缀数组快速找到询问的两个串分别是什么，然后正反各建一个后缀数组来求两个串的LCP和LCS即可。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 typedef long long ll;
 using namespace std;

 const int N=,inf=1e9;
 char s[N];
 ll ans,x,y;
 int n,Q,lg[N];
 struct P{ int x,y; };

 struct SA{
     ll sm[N];
     char s[N];
     int x[N],y[N],c[N],sa[N],rk[N],h[N],mn[N][];

     bool Cmp(int a,int b,int l){ return a+l<=n && b+l<=n && y[a]==y[b] && y[a+l]==y[b+l]; }

     void build(int m){
         rep(i,,m) c[i]=;
         rep(i,,n) c[x[i]=s[i]-'a'+]++;
         rep(i,,m) c[i]+=c[i-];
         for (int i=n; i; i--) sa[c[x[i]]--]=i;
         for (int k=,p=; p<n; k<<=,m=p){
             p=;
             rep(i,n-k+,n) y[++p]=i;
             rep(i,,n) if (sa[i]>k) y[++p]=sa[i]-k;
             rep(i,,m) c[i]=;
             rep(i,,n) c[x[y[i]]]++;
             rep(i,,m) c[i]+=c[i-];
             for (int i=n; i; i--) sa[c[x[y[i]]]--]=y[i];
             rep(i,,n) y[i]=x[i]; x[sa[p=]]=;
             rep(i,,n) x[sa[i]]=Cmp(sa[i],sa[i-],k) ? p : ++p;
         }
     }

     void init(){
         rep(i,,n) rk[sa[i]]=i;
         int k=;
         rep(i,,n){
             for (int j=sa[rk[i]-]; i+k<=n && j+k<=n && s[i+k]==s[j+k]; k++);
             h[rk[i]]=k; if (k) k--;
         }
         rep(i,,n) sm[i]=sm[i-]+n-sa[i]+-h[i],mn[i][]=h[i];
         rep(j,,) rep(i,,n-(<<j)+) mn[i][j]=min(mn[i][j-],mn[i+(<<(j-))][j-]);
     }

     int que(int l,int r){
         if (l==r) return inf;
         l++; int t=lg[r-l+]; return min(mn[l][t],mn[r-(<<t)+][t]);
     }

     P get(ll k){
         int L=,R=n; int ans1,ans2;
         if (sm[n]<k) return (P){-,-};
         while (L<=R){
             int mid=(L+R)>>;
             if (sm[mid]>=k) ans1=mid,ans2=n-sa[mid]+-(sm[mid]-k),R=mid-; else L=mid+;
         }
         return (P){ans1,ans2};
     }
 }S1,S2;

 int main(){
     freopen("bzoj3230.in","r",stdin);
     freopen("bzoj3230.out","w",stdout);
     scanf("%d%d%s",&n,&Q,s+);
     rep(i,,n) lg[i]=lg[i>>]+;
     rep(i,,n) S1.s[i]=S2.s[n-i+]=s[i];
     S1.build(); S2.build(); S1.init(); S2.init();
     rep(i,,Q){
         scanf("%lld%lld",&x,&y); ans=;
         if (S1.sm[n]<y){ puts("-1"); continue; }
         P xx=S1.get(x),yy=S1.get(y);
         ll k=min(S1.que(xx.x,yy.x),min(xx.y,yy.y)); ans+=k*k;
         xx.x=S2.rk[n-S1.sa[xx.x]+-xx.y]; yy.x=S2.rk[n-S1.sa[yy.x]+-yy.y];
         k=min(S2.que(min(xx.x,yy.x),max(xx.x,yy.x)),min(xx.y,yy.y)); ans+=k*k;
         printf("%lld\n",ans);
     }
     return ;
 }