# BZOJ 2648/2716(SJY把件-KD_Tree)[Template:KD_Tree]

## 2648: SJY把件

Time Limit: 20 Sec  Memory Limit: 128 MB
Submit: 1180  Solved: 391
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2 3
1 1
2 3
2 1 2
1 3 3
2 4 2

1
2

kdtree能够过

## Source

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (1000000000)
#define F (100000007)
#define MAXN (500000+10)
#define MAXM (500000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;

int cmp_d=0;
class node
{
public:
int x[2];
int l,r,minv[2],maxv[2];

node(){}
node(int a,int b){MEM(x) l=r=0; x[0]=a,x[1]=b; Rep(i,2) minv[i]=maxv[i]=x[i];}

int& operator[](int i){return x[i];	}
};

int dis(node a,node b){
int ans=0;
Rep(i,2) ans+=abs(a.x[i]-b.x[i]);
return ans;
}

int dis2(node p,node a) // 点p和方形区域a的欧几里德距离
{
int ans=0;
Rep(i,2)
{
if (p.x[i]<a.minv[i]) ans+=a.minv[i]-p.x[i];
else
if (p.x[i]>a.maxv[i]) ans+=p.x[i]-a.maxv[i];
}
return ans;
}

int cmp(node a,node b){return a[cmp_d]<b[cmp_d];	}

class KD_Tree
{
public:
node a[MAXN*3];

KD_Tree()
{
}

void mem()
{
}

void update(node& o)
{
if (o.l)
{
node p=a[o.l];
Rep(i,2) o.minv[i]=min(o.minv[i],p.minv[i]);
Rep(i,2) o.maxv[i]=max(o.maxv[i],p.maxv[i]);
}
if (o.r)
{
node p=a[o.r];
Rep(i,2) o.minv[i]=min(o.minv[i],p.minv[i]);
Rep(i,2) o.maxv[i]=max(o.maxv[i],p.maxv[i]);
}

}

int build(int L,int R,int nowd)
{
int m=(L+R)>>1;

cmp_d=nowd;
nth_element(a+L+1,a+m+1,a+R+1,cmp);

if (L^m) a[m].l=build(L,m-1,nowd^1);
if (R^m) a[m].r=build(m+1,R,nowd^1);

update(a[m]);

return m;

}

int root;
void _build(int L,int R,int nowd)
{
root=build(L,R,nowd);
}

void insert(int o,int k,int nowd)
{
int p=a[o].x[nowd];
int p2=a[k].x[nowd];

if (p2<=p)
{
if (a[o].l)
insert(a[o].l,k,nowd^1);
else a[o].l=k;
}
else
{
if (a[o].r)
insert(a[o].r,k,nowd^1);
else a[o].r=k;

}

update(a[o]);

}
void _insert(int k,int nowd)
{
int p=root;
insert(root,k,nowd);
}

node _p;
int _ans;

{
if (o==0) return;
_ans=min(_ans,dis(a[o],_p));

int ans1=a[o].l ?

dis2(_p,a[a[o].l]) : INF; // 点p到区域内随意一点的距离的最小值 int ans2=a[o].r ? dis2(_p,a[a[o].r]) : INF; if (ans1<ans2) { if(ans1<_ans) ask_min_dis(a[o].l); if(ans2<_ans) ask_min_dis(a[o].r); } else { if(ans2<_ans) ask_min_dis(a[o].r); if(ans1<_ans) ask_min_dis(a[o].l); } } int _ask(node p) { _p=p;_ans=INF; ask_min_dis(root); return _ans; } }S; int main() { // freopen("bzoj2648.in","r",stdin); // freopen("bzoj2648.out","w",stdout); cin>>n>>m; For(i,n) { int x,y; scanf("%d%d",&x,&y); S.a[i]=node(x,y); } S.a[++n]=node(INF,INF); S._build(1,n,0); For(i,m) { int p,x,y; scanf("%d%d%d",&p,&x,&y); if (p==1) { S.a[++n]=node(x,y); S._insert(n,0); } else { printf("%d\n",S._ask(node(x,y))); } } return 0; }

posted @ 2015-10-01 09:59  hrhguanli  阅读(130)  评论(0编辑  收藏  举报