BZOJ4237 : 稻草人

考虑按x坐标排序后分治，只需考虑计算左下角在[l,mid]，右上角在[mid+1,r]的矩形数。

对于[l,mid]的点，从右往左考虑，求出它可以贡献到的纵坐标区间。

对于[mid+1,r]的点，从左往右考虑，求出它可以接受的纵坐标区间。

然后扫描线+Treap维护即可，时间复杂度$O(n\log^2n)$。

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define N 100010
using std::sort;
int n;long long ans;
struct P{int x,y;P(){}P(int _x,int _y){x=_x,y=_y;}}a[N<<1],b
,c
;
inline bool cmp(const P&a,const P&b){return a.x<b.x;}
inline bool cmp2(const P&a,const P&b){return a.y>b.y;}
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
struct node{
int val,sum,p;node*l,*r;
node(){val=sum=p=0;l=r=NULL;}
inline void up(){sum=l->sum+r->sum+1;}
}*blank=new(node),*T,pool[N*3],*cur;
inline void Rotatel(node*&x){node*y=x->r;x->r=y->l;x->up();y->l=x;y->up();x=y;}
inline void Rotater(node*&x){node*y=x->l;x->l=y->r;x->up();y->r=x;y->up();x=y;}
void Ins(node*&x,int y){
if(x==blank){x=cur++;x->val=y;x->l=x->r=blank;x->sum=1;x->p=std::rand();return;}
x->sum++;
if(y<x->val){
Ins(x->l,y);
if(x->l->p>x->p)Rotater(x);
}else{
Ins(x->r,y);
if(x->r->p>x->p)Rotatel(x);
}
}
inline int Pre(node*x,int y){
int t=-1;
while(x!=blank)if(y<x->val)x=x->l;else t=x->val,x=x->r;
return t+1;
}
inline int Nxt(node*x,int y){
int t=1000000001;
while(x!=blank)if(y<x->val)t=x->val,x=x->l;else x=x->r;
return t-1;
}
inline int Ask(node*x,int y){
int t=0;
while(x!=blank)if(y<x->val)x=x->l;else t+=x->l->sum+1,x=x->r;
return t;
}
void solve(int l,int r){
if(l==r)return;
int mid=(l+r)>>1,i,j,m1=0,m2=0;
solve(l,mid),solve(mid+1,r);
for(cur=pool,T=blank,i=mid;i>=l;i--)b[m1++]=P(a[i].y,Nxt(T,a[i].y)),Ins(T,a[i].y);
for(T=blank,i=mid+1;i<=r;i++)c[m2++]=P(Pre(T,a[i].y),a[i].y),Ins(T,a[i].y);
sort(b,b+m1,cmp2),sort(c,c+m2,cmp2);
for(T=blank,i=j=0;i<m2;i++){
while(j<m1&&b[j].y>=c[i].y)Ins(T,b[j++].x);
ans+=Ask(T,c[i].y)-Ask(T,c[i].x-1);
}
}
int main(){
blank->l=blank->r=blank;
read(n);
for(int i=1;i<=n;i++)read(a[i].x),read(a[i].y);
sort(a+1,a+n+1,cmp);
solve(1,n);
return printf("%lld",ans),0;
}