BZOJ 2716 [Violet 3]天使玩偶 ——KD-Tree

【题目分析】

KD-Tree的例题。同BZOJ2648.

【代码】

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>

#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <vector>
#include <iostream>
#include <queue>

using namespace std;

#define maxn 1000005
#define inf 1e9

int read()
{
    int x=0,f=1; char ch=getchar();
    while (ch<'0'||ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0'&&ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}

struct node{
    int mn[2],mx[2],d[2],l,r;
    int operator [] (int x) {return d[x];}
    void init() {d[0]=read(); d[1]=read();}
}t[maxn],now;

int root,D,n,m,tot=0,ans;

int dis(node a,node b){return abs(a[1]-b[1])+abs(a[0]-b[0]);}

bool operator < (node a,node b){return a[D]<b[D]||(a[D]==b[D]&&a[D^1]<b[D^1]);}

void update(int k)
{
    for (int i=0;i<2;++i)
    {
        t[k].mn[i]=min(t[k][i],min(t[t[k].l].mn[i],t[t[k].r].mn[i]));
        t[k].mx[i]=max(t[k][i],max(t[t[k].l].mx[i],t[t[k].r].mx[i]));
    }
}

int build(int l,int r,int dir)
{
    int mid=(l+r)/2;
    D=dir;
    nth_element(t+l,t+mid,t+r+1);
    for (int i=0;i<2;++i) t[mid].mn[i]=t[mid].mx[i]=t[mid][i];
    if (l<mid) t[mid].l=build(l,mid-1,dir^1);
    if (r>mid) t[mid].r=build(mid+1,r,dir^1);
    update(mid);
    return mid;
}

void ins(int & k,int dir)
{
    if (!k)
    {
        k=++tot;
        t[k]=now;
        for (int i=0;i<2;++i) t[k].mn[i]=t[k].mx[i]=t[k][i];
        return ;
    }
    if (now[dir]<t[k][dir]||(now[dir]==t[k][dir]&&now[dir^1]<t[k][dir^1])) ins(t[k].l,dir^1);
    else ins(t[k].r,dir^1);
    update(k);
}

int get(int k)
{
    if (!k) return inf;
    int ret=0;
    for (int i=0;i<2;++i) ret+=max(0,t[k].mn[i]-now[i]);
    for (int i=0;i<2;++i) ret+=max(0,now[i]-t[k].mx[i]);
    return ret;
}

void query(int k)
{
    int dl=get(t[k].l),dr=get(t[k].r),d0=dis(t[k],now);
    ans=min(ans,d0);
    if (dl<dr)
    {
        if (dl<ans) query(t[k].l);
        if (dr<ans) query(t[k].r);
    }
    else
    {
        if (dr<ans) query(t[k].r);
        if (dl<ans) query(t[k].l);
    }
}

int main()
{
    for (int i=0;i<2;++i) t[0].mn[i]=inf,t[0].mx[i]=-inf;
    n=read();m=read();tot=n;
    for (int i=1;i<=n;++i) t[i].init();
    root=build(1,n,0);
    while (m--)
    {
        int opt=read();
        now.init();
        if (opt==1) ins(root,0);
        else
        {
            ans=inf;
            query(root);
            printf("%d\n",ans);
        }
    }
}