[基本操作] kd 树

概念就不说了吧，网上教程满天飞

学了半天才知道，kd 树实质上只干了两件事情：

1.快速定位一个点 / 矩形

2.有理有据地优化暴力

第一点大概是可以来做二维平面上给点/矩形打标记的问题

第二点大概是平面最远点对？

bzoj1941 Hide and Seek

求每个点除自己以外的最近点和最远点

sol：

kd 树优化暴力，对于暴力，考虑这样一个剪枝：如果一个点在某一维上隔的太远，就不搜比它远的了

把这个东西放到 kd 树上就可以了

#include<bits/stdc++.h>
#define LL long long
using namespace std;
inline int read()
{
    int x = 0,f = 1;char ch = getchar();
    for(;!isdigit(ch);ch = getchar())if(ch == '-') f = -f;
    for(;isdigit(ch);ch = getchar())x = 10 * x + ch - '0';
    return x * f;
}
const int maxn = 500010,inf = 1e9;
#define L ps[x].lc
#define R ps[x].rc 
int n,dem,root;
struct Node
{
    int x[2],lc,rc,mx[2],mn[2];
    bool operator < (const Node &b)const{return x[dem] < b.x[dem];}
    friend int dis(Node a,Node b){return abs(a.x[0] - b.x[0]) + abs(a.x[1] - b.x[1]);}
}ps[maxn];
int qx[maxn],qy[maxn];
void maintain(int x)
{
    for(int i=0;i<2;i++)
    {
        ps[x].mn[i] = ps[x].mx[i] = ps[x].x[i];
        if(L)ps[x].mx[i] = max(ps[x].mx[i],ps[L].mx[i]);
        if(L)ps[x].mn[i] = min(ps[x].mn[i],ps[L].mn[i]);
        if(R)ps[x].mx[i] = max(ps[x].mx[i],ps[R].mx[i]);
        if(R)ps[x].mn[i] = min(ps[x].mn[i],ps[R].mn[i]);
    }
}
void build(int &x,int l,int r,int cur)
{
    if(l > r)return;
    dem = cur;
    int mid = (l + r) >> 1;x = mid;
    nth_element(ps + l,ps + mid,ps + r + 1);
    build(L,l,mid - 1,cur ^ 1);build(R,mid + 1,r,cur ^ 1);
    maintain(x);
}
int calmin(Node p,Node cur)
{
    int ans = 0;
    for(int i=0;i<2;i++)
    {
        ans += max(p.x[i] - cur.mx[i],0);
        ans += max(-(p.x[i] - cur.mn[i]),0);
    }
    return ans;
}
int calmax(Node p,Node cur)
{
    int ans = 0;
    for(int i=0;i<2;i++)
        ans += max(abs(p.x[i] - cur.mx[i]),abs(p.x[i] - cur.mn[i]));
    return ans;
}
int ans;
void querymx(Node cur,int x)
{
    ans = max(ans,dis(cur,ps[x]));
    int dl = -inf,dr = -inf;
    if(L)dl = calmax(cur,ps[L]);if(R)dr = calmax(cur,ps[R]);
    if(dl > dr)
    {
        if(dl > ans)querymx(cur,L);
        if(dr > ans)querymx(cur,R);
    }
    else
    {
        if(dr > ans)querymx(cur,R);
        if(dl > ans)querymx(cur,L);
    }
}
void querymn(Node cur,int x)
{
    int tmp = dis(cur,ps[x]);
    if(tmp)ans = min(ans,tmp);
    int dl = inf,dr = inf;
    if(L)dl = calmin(cur,ps[L]);if(R)dr = calmin(cur,ps[R]);
    if(dl < dr)
    {
        if(dl < ans)querymn(cur,L);
        if(dr < ans)querymn(cur,R);
    }
    else
    {
        if(dr < ans)querymn(cur,R);
        if(dl < ans)querymn(cur,L);
    }
}
int query(int x,int y,int type)
{
    Node cur;
    cur.x[0] = x;cur.x[1] = y;
    if(type == 0)ans = inf,querymn(cur,root);
    else ans = -inf,querymx(cur,root);
    return ans;
}
int main()
{
    n = read();
    for(int i=1;i<=n;i++)
    {
        qx[i] = ps[i].x[0] = read();
        qy[i] = ps[i].x[1] = read();        
    }build(root,1,n,0);
    int ret = 2147483233;
    for(int i=1;i<=n;i++)
    {
        int mn = query(qx[i],qy[i],0),mx = query(qx[i],qy[i],1);
        ret = min(ret,mx - mn);
    }cout<<ret<<endl;
}

bzoj4520 K 远点对

给 n 个点的坐标，求第 k 远的点对距离

sol：

暴力的话，用一个长度只有 k 的优先队列维护前 k 远距离

kd 树的话就是用这个做估价函数，如果当前搜到的点进不了队，就不搜了

#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,k;
namespace KD_Tree
{
    int dim;
    priority_queue<ll,vector<ll>,greater<ll> >q;
    void mdf(ll x){q.push(x);q.pop();}
    struct Point
    {
        ll P[2];
        int ls,rs;
        ll mn[2],mx[2];
    }c[100010];
    bool cmp(Point a,Point b){return a.P[dim]<b.P[dim];}
    ll sqr(ll x) {return x*x;}
    ll calc(Point x,Point y){return sqr(x.P[0]-y.P[0])+sqr(x.P[1]-y.P[1]);}
    void update(int x)
    {
        c[x].mn[0]=c[x].mx[0]=c[x].P[0];
        c[x].mn[1]=c[x].mx[1]=c[x].P[1];
        for(int i=0;i<2;i++)
        {
            if(c[x].ls)
            {
                c[x].mn[i]=min(c[x].mn[i],c[c[x].ls].mn[i]);
                c[x].mx[i]=max(c[x].mx[i],c[c[x].ls].mx[i]);
            }
            if(c[x].rs)
            {
                c[x].mn[i]=min(c[x].mn[i],c[c[x].rs].mn[i]);
                c[x].mx[i]=max(c[x].mx[i],c[c[x].rs].mx[i]);
            }
        }
    }
    int build(int l,int r,int w)
    {
        if(l>r) return 0;
        if(l==r) {update(l);return l;}
        dim=w;int mid=(l+r)>>1;
        nth_element(c+l,c+mid,c+r+1,cmp);
        c[mid].ls=build(l,mid-1,w^1);c[mid].rs=build(mid+1,r,w^1);
        update(mid);
        return mid;
    }
    ll cheque(int x,Point p)
    {
        ll tx=max(sqr(c[x].mn[0]-p.P[0]),sqr(c[x].mx[0]-p.P[0]));
        ll ty=max(sqr(c[x].mn[1]-p.P[1]),sqr(c[x].mx[1]-p.P[1]));
        return tx+ty;
    }
    void qury(int x,Point p)
    {
        mdf(calc(c[x],p));
        ll dl=0,dr=0;
        if(c[x].ls) dl=cheque(c[x].ls,p);
        if(c[x].rs) dr=cheque(c[x].rs,p);
        if(dl>dr)
        {
            if(dl>q.top()) qury(c[x].ls,p);
            if(dr>q.top()) qury(c[x].rs,p);
        }
        else
        {
            if(dr>q.top()) qury(c[x].rs,p);
            if(dl>q.top()) qury(c[x].ls,p);
        }
    }
}
using namespace KD_Tree;
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=2*k;i++) q.push(0);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld%lld",&c[i].P[0],&c[i].P[1]);
    }
    int rt=build(1,n,0);
    for(int i=1;i<=n;i++)
    {
        qury(rt,c[i]);
    }
    printf("%lld
",q.top());
}

bzoj2683 & bzoj4066 简单题

矩形加，矩形求和

sol：

这道题要支持在 kd 树里插入一个点，类似二叉搜索树，走到儿子然后把新点加进去就可以了

注意定期重构

重构的时候注意走到非法节点就把它赋成 0

#include<bits/stdc++.h>
#define LL long long
using namespace std;
inline int read()
{
    int x = 0,f = 1;char ch = getchar();
    for(;!isdigit(ch);ch = getchar())if(ch == '-') f = -f;
    for(;isdigit(ch);ch = getchar())x = 10 * x + ch - '0';
    return x * f;
}
const int maxn = 200010,inf = 1e9;
#define L (ps[x].lc)
#define R (ps[x].rc) 
int n,dem,root,ToT;
struct Node
{
    int x[2],lc,rc,mx[2],mn[2],val,sum;
    bool operator < (const Node &b)const{return x[dem] < b.x[dem];}
    bool operator == (const Node& t) const { return x[0] == t.x[0] && x[1] == t.x[1]; }
    //friend int dis(Node a,Node b){return abs(a.x[0] - b.x[0]) + abs(a.x[1] - b.x[1]);}
}ps[maxn];
void maintain(int x)
{
    for(int i=0;i<2;i++)
    {
        ps[x].mn[i] = ps[x].mx[i] = ps[x].x[i];
        //cout<<L<<" "<<R<<endl;
        if(L)ps[x].mx[i] = max(ps[x].mx[i],ps[L].mx[i]);
        if(L)ps[x].mn[i] = min(ps[x].mn[i],ps[L].mn[i]);
        if(R)ps[x].mx[i] = max(ps[x].mx[i],ps[R].mx[i]);
        if(R)ps[x].mn[i] = min(ps[x].mn[i],ps[R].mn[i]);
    }
    ps[x].sum = ps[x].val;
    if(L)ps[x].sum += ps[L].sum;
    if(R)ps[x].sum += ps[R].sum;
}
void build(int &x,int l,int r,int cur)
{
    if(l > r){x = 0;return;}
    dem = cur;
    int mid = (l + r) >> 1;x = mid;
    nth_element(ps + l,ps + mid,ps + r + 1);
    build(L,l,mid - 1,cur ^ 1);build(R,mid + 1,r,cur ^ 1);
    maintain(x);
}
void insert(int &x,Node cur,int cur_d)
{
    if(!x)
    {
        x = ++ToT;
        ps[x] = cur;
        maintain(x);
        return;
    }
    if(ps[x] == cur)
    {
        ps[x].val += cur.val;
        ps[x].sum += cur.val;
        maintain(x);
        return; 
    }
    if(cur.x[cur_d] < ps[x].x[cur_d])insert(L,cur,cur_d ^ 1);
    else insert(R,cur,cur_d ^ 1);
    maintain(x);
}
inline int all_in(Node cur_1,Node cur_2,int x){return (cur_1.x[0] <= ps[x].mn[0]) && (ps[x].mx[0] <= cur_2.x[0]) && (cur_1.x[1] <= ps[x].mn[1]) && (ps[x].mx[1] <= cur_2.x[1]);}
inline int has_node(Node cur_1,Node cur_2,int x){return !(ps[x].mx[0] < cur_1.x[0] || ps[x].mn[0] > cur_2.x[0] || ps[x].mx[1] < cur_1.x[1] || ps[x].mn[1] > cur_2.x[1]);}
inline int query(Node cur_1,Node cur_2,int x)
{
    if(!x)return 0;
    int ans = 0;
    if(all_in(cur_1,cur_2,L))ans += ps[L].sum;
    else if(has_node(cur_1,cur_2,L)) ans += query(cur_1,cur_2,L);
    if(all_in(cur_1,cur_2,R))ans += ps[R].sum;
    else if(has_node(cur_1,cur_2,R)) ans += query(cur_1,cur_2,R);
    int nx = ps[x].x[0],ny = ps[x].x[1];
    if(cur_1.x[0] <= nx && nx <= cur_2.x[0] && cur_1.x[1] <= ny && ny <= cur_2.x[1])ans += ps[x].val;
    return ans;
}Node cur_1,cur_2; 
int main()
{
    //nodes[0].val = nodes[0].sum = 0;
    n = read();int lastans = 0;
    n = read();
    while(n != 3)
    {
        if(n == 1)
        {
            cur_1.x[0] = read() ^ lastans;
            cur_1.x[1] = read() ^ lastans;
            cur_1.val = read() ^ lastans;
            insert(root,cur_1,1);if(ToT % 1000 == 0)build(root,1,ToT,1);
        }
        if(n == 2)
        {
            cur_1.x[0] = read() ^ lastans;cur_1.x[1] = read() ^ lastans;
            cur_2.x[0] = read() ^ lastans;cur_2.x[1] = read() ^ lastans;
            printf("%d
",lastans = query(cur_1,cur_2,root));
        }
        n = read();
    }
}

bzoj4170 & 2989 数列

给一个二维平面，每次插入一个单点，查询与给定点曼哈顿距离不超过 $d$ 的点的数量

sol：

实质上是查询一个斜 45 度的正方形内有多少个点，这个东西直接做复杂度好像是错的，旋转一下坐标系即可

（然而你为啥不写 CDQ 呢

#include<bits/stdc++.h>
#define LL long long
using namespace std;
inline int read()
{
    int x = 0,f = 1;char ch = getchar();
    for(;!isdigit(ch);ch = getchar())if(ch == '-') f = -f;
    for(;isdigit(ch);ch = getchar())x = 10 * x + ch - '0';
    return x * f;
}
const int maxn = 150010;
int n,q,dem,ToT,root;
int a[maxn];
#define L tr[x].ls
#define R tr[x].rs 
struct Node
{
    int x[2],ls,rs,mx[2],mn[2],val,sum;
    bool operator < (const Node &b)const{return x[dem] < b.x[dem];}
    bool operator == (const Node &b)const{return (x[1] == b.x[1]) && (x[0] == b.x[0]);}
}tr[maxn];
void maintain(int x)
{
    for(int i=0;i<2;i++)
    {
        tr[x].mx[i] = tr[x].mn[i] = tr[x].x[i];
        if(L)tr[x].mx[i] = max(tr[L].mx[i],tr[x].mx[i]);
        if(L)tr[x].mn[i] = min(tr[L].mn[i],tr[x].mn[i]);
        if(R)tr[x].mx[i] = max(tr[R].mx[i],tr[x].mx[i]);
        if(R)tr[x].mn[i] = min(tr[R].mn[i],tr[x].mn[i]); 
    }tr[x].sum = tr[x].val;
    if(L)tr[x].sum += tr[L].sum;
    if(R)tr[x].sum += tr[R].sum;
}
void build(int &x,int l,int r,int cur_d)
{
    if(l > r){x = 0;return;}
    dem = cur_d;
    int mid = (l + r) >> 1;
    nth_element(tr + l,tr + mid,tr + r + 1);
    x = mid;//tr[x].val = 1;
    cur_d = cur_d ^ 1;
    build(L,l,mid - 1,cur_d);build(R,mid + 1,r,cur_d);
    maintain(x);
}
void insert(int &x,Node cur,int cur_d)
{
    if(!x)
    {
        x = ++ToT;
        tr[x] = cur;
        maintain(x);
        return;
    }
    if(tr[x] == cur)
    {
        tr[x].val += cur.val;
        tr[x].sum += cur.val;
        maintain(x);
        return; 
    }
    if(cur.x[cur_d] < tr[x].x[cur_d])insert(L,cur,cur_d ^ 1);
    else insert(R,cur,cur_d ^ 1);
    maintain(x);
}
inline int all_in(Node cur_1,Node cur_2,int x){return (cur_1.x[0] <= tr[x].mn[0]) && (tr[x].mx[0] <= cur_2.x[0]) && (cur_1.x[1] <= tr[x].mn[1]) && (tr[x].mx[1] <= cur_2.x[1]);}
inline int has_node(Node cur_1,Node cur_2,int x){return !(tr[x].mx[0] < cur_1.x[0] || tr[x].mn[0] > cur_2.x[0] || tr[x].mx[1] < cur_1.x[1] || tr[x].mn[1] > cur_2.x[1]);}
inline int query(Node cur_1,Node cur_2,int x)
{
    if(!x)return 0;
    int ans = 0;
    if(all_in(cur_1,cur_2,L))ans += tr[L].sum;
    else if(has_node(cur_1,cur_2,L)) ans += query(cur_1,cur_2,L);
    if(all_in(cur_1,cur_2,R))ans += tr[R].sum;
    else if(has_node(cur_1,cur_2,R)) ans += query(cur_1,cur_2,R);
    int nx = tr[x].x[0],ny = tr[x].x[1];
    if(cur_1.x[0] <= nx && nx <= cur_2.x[0] && cur_1.x[1] <= ny && ny <= cur_2.x[1])ans += tr[x].val;
    return ans;
}Node cur_1,cur_2;
int main()
{
    n = read();q = read();
    for(int i=1;i<=n;i++)a[i] = read();
    for(int i=1;i<=n;i++)
    {
        cur_1.x[0] = i - a[i],cur_1.x[1] = i + a[i],cur_1.val = 1;
        insert(root,cur_1,1);
    }build(root,1,n,1);
    char opt[50];
    while(q--)
    {
        scanf("%s",opt + 1);
        if(opt[1] == 'M')
        {
            int x = read(),y = read();
            a[x] = y;cur_1.x[0] = x - y,cur_1.x[1] = x + y,cur_1.val = 1;
            insert(root,cur_1,1);if(ToT % 2000 == 0)build(root,1,ToT,1);
        }
        else
        {
            int x = read(),y = read();
            cur_1.x[0] = x - a[x] - y;
            cur_2.x[1] = x + a[x] + y;
            cur_2.x[0] = x - a[x] + y;
            cur_1.x[1] = x + a[x] - y;
            //if(cur_1.x[0] > cur_2.x[0])swap(cur_1.x[0],cur_2.x[0]);
            //if(cur_1.x[1] > cur_2.x[1])swap(cur_1.x[1],cur_2.x[1]);
            printf("%d
",query(cur_1,cur_2,root));
        }
    }
}