题解：
傻逼题
直接从左向右扫描每个点作为右端点
然后单点修改区间查询就行了
另外一种更直观的做法就是$(i,j)$之间产生了$(j-i)$
于是变成矩形查最大值，kd-tree维护
代码：
#include <bits/stdc++.h> #define rint register int #define IL inline #define rep(i,h,t) for (int i=h;i<=t;i++) #define dep(i,t,h) for (int i=t;i>=h;i--) #define me(x) memset(x,0,sizeof(x)) #define ll long long namespace IO{ char ss[1<<24],*A=ss,*B=ss; IL char gc() { return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++; } template<class T>void read(T &x) { rint f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48); while (c=gc(),c>47&&c<58) x=(x<<3)+(x<<1)+(c^48); x*=f; } char sr[1<<24],z[20]; int Z,C=-1; template<class T>void wer(T x) { if (x<0) sr[++C]='-',x=-x; while (z[++Z]=x%10+48,x/=10); while (sr[++C]=z[Z],--Z); } IL void wer1() {sr[++C]=' ';} IL void wer2() {sr[++C]=' ';} template<class T>IL void maxa(T &x,T y) {if (x<y) x=y;} template<class T>IL void mina(T &x,T y) {if (x>y) x=y;} template<class T>IL T MAX(T x,T y) { return x>y?x:y;} template<class T>IL T MIN(T x,T y) { return x<y?x:y;} }; using namespace std; using namespace IO; const int N=6e5; map<int,int> M; int cnt,a[N],cmp_d,rt; struct re{ int x,y,z; }p[N]; bool cmp(re x,re y) { if (!cmp_d) return x.x<y.x; else return x.y<y.y; } const int INF=1e9; struct kd_tree{ int v[N],pv[N],ls[N],rs[N],Mx[N],Nx[N],My[N],Ny[N]; kd_tree() { v[0]=pv[0]=INF; Nx[0]=INF; Mx[0]=0; Ny[0]=INF; My[0]=0; } IL void updata(int x) { pv[x]=MIN(MIN(pv[ls[x]],pv[rs[x]]),v[x]); Mx[x]=MAX(p[x].x,MAX(Mx[ls[x]],Mx[rs[x]])); Nx[x]=MIN(p[x].x,MIN(Nx[ls[x]],Nx[rs[x]])); My[x]=MAX(p[x].y,MAX(My[ls[x]],My[rs[x]])); Ny[x]=MIN(p[x].y,MIN(Ny[ls[x]],Ny[rs[x]])); } int build(int h,int t,int o) { int x,mid; x=mid=(h+t)/2; cmp_d=o; nth_element(p+h,p+mid,p+t+1,cmp); v[x]=p[x].z; if (h<mid) ls[x]=build(h,mid-1,o^1); if (mid<t) rs[x]=build(mid+1,t,o^1); updata(x); return x; } int query(int x,int x1,int x2,int y1,int y2) { if (x1>Mx[x]||x2<Nx[x]||y1>My[x]||y2<Ny[x]) return(INF); if (x1<=Nx[x]&&Mx[x]<=x2&&y1<=Ny[x]&&My[x]<=y2) return(pv[x]); int ans=INF; if (p[x].x>=x1&&p[x].x<=x2&&p[x].y>=y1&&p[x].y<=y2) ans=v[x]; mina(ans,query(ls[x],x1,x2,y1,y2)); mina(ans,query(rs[x],x1,x2,y1,y2)); return ans; } }K; int main() { freopen("1.in","r",stdin); freopen("1.out","w",stdout); int n,m; read(n); read(m); rep(i,1,n) { read(a[i]); int k=M[a[i]]; if (k) p[++cnt]=(re){k,i,i-k}; M[a[i]]=i; } rt=K.build(1,cnt,0); rep(i,1,m) { int x,y; read(x); read(y); int ans=K.query(rt,x,y,x,y); if (ans<INF) wer(ans); else wer(-1); wer2(); } fwrite(sr,1,C+1,stdout); return 0; }