Codeforces 522D Closest Equals

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]=‘\n‘;}
    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;
}

原文地址：https://www.cnblogs.com/yinwuxiao/p/10100457.html