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$O(nlog^2n)$。
#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[N],c[N]; 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; }