题解:
并没有想到。。
我们用前驱后继来表示出题目要求
即pre[i]<l且scc[i]>r且l<=i<=r
即4维偏序
使用树套树可以用主席树套线段树(很毒瘤??)
那么会kd-tree就比较简单了,3维空间查找
代码写完编译运行一遍过就很爽了
代码:
#include <bits/stdc++.h> using namespace std; #define IL inline #define rint register int #define rep(i,h,t) for (rint i=h;i<=t;i++) #define dep(i,t,h) for (rint i=t;i>=h;i--) #define mid ((h+t)/2) const int INF=1e9; const int N=3e5; int n,m; 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; } IL void umax(int &x,int y) { if (x<y) x=y; } IL void umin(int &x,int y) { if (x>y) x=y; } struct re { int d[3],num; }p[N]; int cmp_d,ans,a[N]; bool cmp(re x,re y) { return(x.d[cmp_d]<y.d[cmp_d]); } struct kd { int ls[N],rs[N],Mx[N],My[N],Mz[N],Nx[N],Ny[N],Nz[N],maxn[N]; void updata(int x) { if (ls[x]) { umax(Mx[x],Mx[ls[x]]); umax(My[x],My[ls[x]]); umax(Mz[x],Mz[ls[x]]); umin(Nx[x],Nx[ls[x]]); umin(Ny[x],Ny[ls[x]]); umin(Nz[x],Nz[ls[x]]); umax(maxn[x],maxn[ls[x]]); } if (rs[x]) { umax(Mx[x],Mx[rs[x]]); umax(My[x],My[rs[x]]); umax(Mz[x],Mz[rs[x]]); umin(Nx[x],Nx[rs[x]]); umin(Ny[x],Ny[rs[x]]); umin(Nz[x],Nz[rs[x]]); umax(maxn[x],maxn[rs[x]]); } } int build(int h,int t,int o) { cmp_d=o; nth_element(p+h,p+mid,p+t+1,cmp); int x=mid; Mx[x]=Nx[x]=p[x].d[0]; My[x]=Ny[x]=p[x].d[1]; Mz[x]=Nz[x]=p[x].d[2]; maxn[x]=p[x].num; if (h!=x) ls[x]=build(h,mid-1,(o+1)%3); else ls[x]=0; if (t!=x) rs[x]=build(mid+1,t,(o+1)%3); else rs[x]=0; updata(x); return x; } void query(int k,int x1,int x2,int y,int z) { if (!k||maxn[k]<ans||Mx[k]<x1||Nx[k]>x2||Ny[k]>=y||Mz[k]<=z) return; if (p[k].d[0]>=x1&&p[k].d[0]<=x2&&p[k].d[1]<y&&p[k].d[2]>z&&p[k].num>ans) ans=p[k].num; query(ls[k],x1,x2,y,z); query(rs[k],x1,x2,y,z); } }kd; map<int,int> M; int pre[N],scc[N]; int main() { freopen("1.in","r",stdin); freopen("1.out","w",stdout); read(n); read(m); rep(i,1,n) { read(a[i]); int kk=M[a[i]]; if (kk) { scc[kk]=i; pre[i]=kk; } M[a[i]]=i; } rep(i,1,n) if (!scc[i]) scc[i]=n+1; rep(i,1,n) p[i].num=a[i],p[i].d[0]=i,p[i].d[1]=pre[i],p[i].d[2]=scc[i]; int rt=kd.build(1,n,0); rep(i,1,m) { int x,y; read(x); read(y); x=(ans+x)%n+1,y=(ans+y)%n+1; if (x>y) swap(x,y); ans=0; kd.query(rt,x,y,x,y); printf("%d ",ans); } return 0; }