[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=100010,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][20];
    
    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,0,m) c[i]=0;
        rep(i,1,n) c[x[i]=s[i]-'a'+1]++;
        rep(i,1,m) c[i]+=c[i-1];
        for (int i=n; i; i--) sa[c[x[i]]--]=i;
        for (int k=1,p=0; p<n; k<<=1,m=p){
            p=0;
            rep(i,n-k+1,n) y[++p]=i;
            rep(i,1,n) if (sa[i]>k) y[++p]=sa[i]-k;
            rep(i,0,m) c[i]=0;
            rep(i,1,n) c[x[y[i]]]++;
            rep(i,1,m) c[i]+=c[i-1];
            for (int i=n; i; i--) sa[c[x[y[i]]]--]=y[i];
            rep(i,1,n) y[i]=x[i]; x[sa[p=1]]=1;
            rep(i,2,n) x[sa[i]]=Cmp(sa[i],sa[i-1],k) ? p : ++p;
        }
    }
    
    void init(){
        rep(i,1,n) rk[sa[i]]=i;
        int k=0;
        rep(i,1,n){
            for (int j=sa[rk[i]-1]; i+k<=n && j+k<=n && s[i+k]==s[j+k]; k++);
            h[rk[i]]=k; if (k) k--;
        }
        rep(i,1,n) sm[i]=sm[i-1]+n-sa[i]+1-h[i],mn[i][0]=h[i];
        rep(j,1,18) rep(i,1,n-(1<<j)+1) mn[i][j]=min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]);
    }
    
    int que(int l,int r){
        if (l==r) return inf;
        l++; int t=lg[r-l+1]; return min(mn[l][t],mn[r-(1<<t)+1][t]);
    }
    
    P get(ll k){
        int L=1,R=n; int ans1,ans2;
        if (sm[n]<k) return (P){-1,-1};
        while (L<=R){
            int mid=(L+R)>>1;
            if (sm[mid]>=k) ans1=mid,ans2=n-sa[mid]+1-(sm[mid]-k),R=mid-1; else L=mid+1;
        }
        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+1);
    rep(i,2,n) lg[i]=lg[i>>1]+1;
    rep(i,1,n) S1.s[i]=S2.s[n-i+1]=s[i];
    S1.build(30); S2.build(30); S1.init(); S2.init();
    rep(i,1,Q){
        scanf("%lld%lld",&x,&y); ans=0;
        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]+2-xx.y]; yy.x=S2.rk[n-S1.sa[yy.x]+2-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
",ans);
    }
    return 0;
}