bzoj 4066 & bzoj 2683 简单题 —— K-D树(含重构)

题目:https://www.lydsy.com/JudgeOnline/problem.php?id=4066

https://www.lydsy.com/JudgeOnline/problem.php?id=2683

高仿:https://www.cnblogs.com/Narh/p/9605505.html

注意细节...

AC 300 ~

代码如下:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
int const xn=2e5+5;
double const alpha=0.75;
int n,cnt,sta[xn],top,rt,dm,c[xn][2],qx1,qx2,qy1,qy2;
int R,fa,son,num;
ll ans;
struct N{
  int x[2],y[2],p[2],siz;
  ll sum,ys;
}t[xn],a[xn];
int rd()
{
  int ret=0,f=1; char ch=getchar();
  while(ch<'0'||ch>'9'){if(ch=='-')f=0; ch=getchar();}
  while(ch>='0'&&ch<='9')ret=ret*10+ch-'0',ch=getchar();
  return f?ret:-ret;
}
ll Abs(ll x){return x>0?x:-x;}
ll Min(ll x,ll y){return x<y?x:y;}
ll Max(ll x,ll y){return x<y?y:x;}
bool cmp(N x,N y){return x.p[dm]<y.p[dm];}
int node(){if(top)return sta[top--]; else return ++cnt;}
void turn(int x,N v)
{
  for(int i=0;i<2;i++)t[x].x[i]=t[x].y[i]=t[x].p[i]=v.p[i]; 
  t[x].ys=t[x].sum=v.ys; t[x].siz=1; c[x][0]=0; c[x][1]=0;//ys //c //v.ys!!!
}
void pushup(int x)
{
  int ls=c[x][0],rs=c[x][1];
  for(int i=0;i<2;i++)
    {
      if(ls)t[x].x[i]=Min(t[x].x[i],t[ls].x[i]),t[x].y[i]=Max(t[x].y[i],t[ls].y[i]);
      if(rs)t[x].x[i]=Min(t[x].x[i],t[rs].x[i]),t[x].y[i]=Max(t[x].y[i],t[rs].y[i]);
    }
  t[x].sum=t[x].ys+(ls?t[ls].sum:0)+(rs?t[rs].sum:0);
  t[x].siz=1+(ls?t[ls].siz:0)+(rs?t[rs].siz:0);
}
bool ck(int x)
{
  int ls=c[x][0],rs=c[x][1];
  return (ls&&t[ls].siz>t[x].siz*alpha)||(rs&&t[rs].siz>t[x].siz*alpha);
}
void ins(int &x,N v,int nw)
{
  if(!x){x=node(); turn(x,v); return;}
  if(v.p[nw]<=t[x].p[nw])ins(c[x][0],v,nw^1);
  else ins(c[x][1],v,nw^1);
  pushup(x);
  if(ck(x))R=x,dm=nw;
  if(R==c[x][0])fa=x,son=0; if(R==c[x][1])fa=x,son=1;
}
bool can(int x)
{return t[x].p[0]>=qx1&&t[x].p[0]<=qx2&&t[x].p[1]>=qy1&&t[x].p[1]<=qy2;}
bool in(int x)
{return t[x].x[0]>=qx1&&t[x].y[0]<=qx2&&t[x].x[1]>=qy1&&t[x].y[1]<=qy2;}
bool out(int x)
{return t[x].x[0]>qx2||t[x].y[0]<qx1||t[x].x[1]>qy2||t[x].y[1]<qy1;}
ll query(int x)
{
  if(can(x))ans+=t[x].ys;//ys
  int ls=c[x][0],rs=c[x][1];
  if(ls)
    {
      if(in(ls))ans+=t[ls].sum;
      else if(!out(ls))query(ls);
    }
  if(rs)
    {
      if(in(rs))ans+=t[rs].sum;
      else if(!out(rs))query(rs);
    }
}
void kill(int x)
{
  sta[++top]=x; a[++num]=t[x];
  if(c[x][0])kill(c[x][0]);
  if(c[x][1])kill(c[x][1]);
}
void build(int &x,int l,int r,int nw)
{
  x=node(); dm=nw; int mid=((l+r)>>1);
  nth_element(a+l,a+mid,a+r+1,cmp);
  turn(x,a[mid]);
  if(mid>l)build(c[x][0],l,mid-1,nw^1);
  if(mid<r)build(c[x][1],mid+1,r,nw^1);
  pushup(x);
}
void rbuild(int x)
{
  num=0; kill(x); int mid=((1+num)>>1);
  nth_element(a+1,a+mid,a+num+1,cmp);
  int p=node(); turn(p,a[mid]);
  if(fa)c[fa][son]=p; else rt=p;//
  if(mid>1)build(c[p][0],1,mid-1,dm^1);
  if(mid<num)build(c[p][1],mid+1,num,dm^1);
  pushup(p);
}
int main()
{
  n=rd(); N tmp;
  while((n=rd())!=3)
    {
      if(n==1)
	{
	  //tmp.p[0]=(rd()^ans); tmp.p[1]=(rd()^ans); tmp.ys=(rd()^ans);
	  tmp.p[0]=rd(); tmp.p[1]=rd(); tmp.ys=rd();
	  fa=0; ins(rt,tmp,0);//fa
	}
      else
	{
	  //qx1=(rd()^ans); qy1=(rd()^ans);
	  //qx2=(rd()^ans); qy2=(rd()^ans);
	  qx1=rd(); qy1=rd(); qx2=rd(); qy2=rd();
	  ans=0; query(rt); printf("%lld\n",ans);
	}
      if(R){if(R==rt)fa=0; rbuild(R); R=0;}
    }
  return 0;
}

 

posted @ 2018-12-21 00:28  Zinn  阅读(162)  评论(0编辑  收藏  举报