传送门
后缀数组入门题。
建立正反两个后缀数组算就行了。
代码：
#include<bits/stdc++.h>
#define ri register int
using namespace std;
typedef long long ll;
const int N=2e5+5;
int n,m,q,sa2[N],Log[N],cnt[N];
ll num[N];
char s[N];
struct SA{
int sa[N],rk[N],ht[N],st[N][21];
inline void Sort(){
for(ri i=1;i<=m;++i)cnt[i]=0;
for(ri i=1;i<=n;++i)++cnt[rk[i]];
for(ri i=2;i<=m;++i)cnt[i]+=cnt[i-1];
for(ri i=n;i;--i)sa[cnt[rk[sa2[i]]]--]=sa2[i];
}
inline void getsa(){
for(ri i=1;i<=n;++i)rk[i]=s[i]-'a'+1,sa2[i]=i;
m=26,Sort();
for(ri w=1,p=0;m^n;w<<=1,p=0){
for(ri i=n-w+1;i<=n;++i)sa2[++p]=i;
for(ri i=1;i<=n;++i)if(sa[i]>w)sa2[++p]=sa[i]-w;
Sort(),swap(sa2,rk),rk[sa[1]]=p=1;
for(ri i=2;i<=n;++i)rk[sa[i]]=(sa2[sa[i]]==sa2[sa[i-1]]&&sa2[sa[i]+w]==sa2[sa[i-1]+w])?p:++p;
m=p;
}
for(ri i=1,j,k=0;i<=n;ht[rk[i++]]=k)for(k?--k:k,j=sa[rk[i]-1];s[i+k]==s[j+k];++k);
for(ri i=1;i<=n;++i)st[i][0]=ht[i];
for(ri j=1;j<=20;++j)for(ri i=1;i+(1<<j)-1<=n;++i)st[i][j]=min(st[i][j-1],st[i+(1<<(j-1))][j-1]);
}
inline int rmq(int l,int r){return min(st[l][Log[r-l+1]],st[r-(1<<Log[r-l+1])+1][Log[r-l+1]]);}
inline int query(int l,int r){
l=rk[l],r=rk[r];
if(l>r)l^=r,r^=l,l^=r;
return rmq(l+1,r);
}
}A,B;
inline ll read(){
ll ans=0;
char ch=getchar();
while(!isdigit(ch))ch=getchar();
while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
return ans;
}
inline void solve(int&l,int&r,ll k){
int id=lower_bound(num+1,num+n+1,k)-num;
l=A.sa[id],r=A.sa[id]+A.ht[id]+k-num[id-1]-1;
}
int main(){
freopen("lx.in","r",stdin);
n=read(),q=read(),scanf("%s",s+1),Log[0]=-1,A.getsa(),reverse(s+1,s+n+1),B.getsa();
for(ri i=1;i<=n;++i)Log[i]=Log[i>>1]+1,num[i]=num[i-1]+(ll)(n-A.sa[i]+1-A.ht[i]);
while(q--){
ll a=read(),b=read(),sum=0,ans;
if(a>num[n]||b>num[n]){puts("-1");continue;}
int l1,r1,l2,r2;
solve(l1,r1,a),solve(l2,r2,b);
ans=l1==l2?0x3f3f3f3f:A.query(l1,l2);
ans=min(ans,(ll)min(r1-l1+1,r2-l2+1));
sum+=ans*ans;
ans=r1==r2?0x3f3f3f3f:B.query(n-r1+1,n-r2+1);
ans=min(ans,(ll)min(r1-l1+1,r2-l2+1));
sum+=ans*ans;
cout<<sum<<'\n';
}
return 0;
}