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);
}