BZOJ 1818 线段树+扫描线

思路：
可以把题目转化成
给你一些沿坐标轴方向的线段 让你求交点个数
然后就线段树+扫描线 搞一搞
（线段不包含断点 最后+n 这种方式 比线段包含断点+各种特判要好写得多）

//By SiriusRen#include <cstdio>#include <algorithm>using namespace std;const int N=100005;int n,stkx[N],stky[N],topx,topy,top,tree[N*4];long long ans;struct Node{int x,y;}node[N];bool cmp1(Node a,Node b){if(a.x!=b.x)return a.x<b.x;return a.y<b.y;}bool cmp2(Node a,Node b){if(a.y!=b.y)return a.y<b.y;return a.x<b.x;}struct Movement{    int time,num,move,l,r;    Movement(int x,int y,int z){time=x,num=y,move=z;}    Movement(int x,int y,int z,int flg){time=x,l=y,r=z,move=flg;}    Movement(){}    friend bool operator<(Movement a,Movement b){        if(a.time!=b.time)return a.time<b.time;        return a.move<b.move;    }}movement[N*3];void insert(int l,int r,int pos,int num,int wei){    if(l==r){tree[pos]+=wei;return;}    int mid=(l+r)>>1,lson=pos<<1,rson=pos<<1|1;    if(mid<num)insert(mid+1,r,rson,num,wei);    else insert(l,mid,lson,num,wei);    tree[pos]=tree[lson]+tree[rson];}int query(int l,int r,int pos,int L,int R){    if(L>R)return 0;    if(l>=L&&r<=R){return tree[pos];}    int mid=(l+r)>>1,lson=pos<<1,rson=pos<<1|1;    if(mid<L)return query(mid+1,r,rson,L,R);    else if(mid>=R)return query(l,mid,lson,L,R);    else return query(l,mid,lson,L,R)+query(mid+1,r,rson,L,R);}signed main(){    scanf("%d",&n);    for(int i=1;i<=n;i++)        scanf("%d%d",&node[i].x,&node[i].y),        stkx[++topx]=node[i].x,stky[++topy]=node[i].y;    sort(stkx+1,stkx+1+topx),sort(stky+1,stky+1+topy);    topx=unique(stkx+1,stkx+1+topx)-stkx-1;    topy=unique(stky+1,stky+1+topy)-stky-1;    for(int i=1;i<=n;i++)        node[i].x=lower_bound(stkx+1,stkx+1+topx,node[i].x)-stkx,         node[i].y=lower_bound(stky+1,stky+1+topy,node[i].y)-stky;    sort(node+1,node+1+n,cmp1);    for(int i=1;i<=n;i++)if(node[i].x==node[i-1].x)        movement[++top]=Movement(node[i-1].y,node[i].x,1),        movement[++top]=Movement(node[i].y,node[i].x,-1);    sort(node+1,node+1+n,cmp2);    for(int i=1;i<=n;i++)if(node[i].y==node[i-1].y)        movement[++top]=Movement(node[i].y,node[i-1].x,node[i].x,0);    sort(movement+1,movement+1+top);    for(int i=1;i<=top;i++)        if(movement[i].move)insert(1,topx,1,movement[i].num,movement[i].move);        else ans+=query(1,topx,1,movement[i].l+1,movement[i].r-1);    printf("%lld\n",ans+n);}