bzoj1935 [Shoi2007] Tree gardener's troubles, two-dimensional partial order

In fact, this question is a two-dimensional partial order, be careful not to think about the tree set tree

Use tolerance and exclusion to find the partial imaginary prefix sum of 4 points

To ensure that y<required points, sort by y and process each query point separately

To ensure that x> the required point, sum the prefix of the tree array

code:

#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
struct dian 
{
int x,y,id,wz;	
}D[2000005],W[8000005];
int v[12000007],n,m,i,xxx[8000005],yyy[8000005],cntx,cnty,cntw,zx,zy,yx,yy,o,daan[2000005];
map<int,int>mx,my;
bool cmp(dian a,dian b)
{
	return a.y<b.y;
}
int lowbit(int o)
{
	return o&(-o);
}
void jia(int o)
{
	for(;o<=12000005;o+=lowbit(o))
	{
		v[o]+=1;
	}	
}
int cha(int o)
{
	int ans=0;
	for(;o>0;o-=lowbit(o))
	{
		ans+=v[o];	
	}
	return ans;	
}
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)
	scanf("%d%d",&D[i].x,&D[i].y),xxx[++cntx]=D[i].x,yyy[++cnty]=D[i].y;
	D[++n].x=999999999;D[n].y=999999999,xxx[++cntx]=D[n].x,yyy[++cnty]=D[n].y;
	for(i=1;i<=m;i++)
	{
		scanf("%d%d%d%d",&zx,&zy,&yx,&yy);
	zx--;zy--;
		W[++cntw].id=i;W[cntw].wz=1;W[cntw].x=zx;W[cntw].y=yy;
		W[++cntw].id=i;W[cntw].wz=2;W[cntw].x=yx;W[cntw].y=yy;
		W[++cntw].id=i;W[cntw].wz=3;W[cntw].x=zx;W[cntw].y=zy;
		W[++cntw].id=i;W[cntw].wz=4;W[cntw].x=yx;W[cntw].y=zy;
		xxx[++cntx]=zx;xxx[++cntx]=yx;
		yyy[++cnty]=zy;yyy[++cnty]=yy;			
	}
	sort(xxx+1,xxx+1+cntx);
	for(i=1;i<=cntx;i++)
	mx[xxx[i]]=i;
	sort(yyy+1,yyy+1+cnty);
	for(i=1;i<=cnty;i++)
	my[yyy[i]]=i;	
	sort(D+1,D+1+n,cmp);
	sort(W+1,W+1+cntw,cmp);
	o=1;
	for(i=1;i<=n;i++)
	{
	while(W[o].y<D[i].y&&o<=cntw)
		{  int lin=cha(mx[W[o].x]);
			if(W[o].wz==1)daan[W[o].id]-=lin;
			if(W[o].wz==2)daan[W[o].id]+=lin;
			if(W[o].wz==3)daan[W[o].id]+=lin;
			if(W[o].wz==4)daan[W[o].id]-=lin;
			o++;
		}
		jia(mx[D[i].x]);		
	}
	for(i=1;i<=m;i++)
	printf("%d\n",daan[i]);
}