传送门：hdu 5862 Counting Intersections
题意：对于平行于坐标轴的n条线段，求两两相交的线段对有多少个，包括十，T型
官方题解：由于数据限制,只有竖向与横向的线段才会产生交点,所以先对横向线段按x端点排序,每次加入一个线段,将其对应的y坐标位置+1,当出现一个竖向线段时,查询它的两个y端点之间的和即为交点个数.
注意点：对x坐标排序是对所有线段端点排序；因为可能出现 “ 1-1 “ 这样的情况，所以对于横着的线段，需要进行首尾x坐标处理；我的方法是对于x坐标，先按大小排序，如果大小相同，按先横后竖的方式排序，那么对于 “ 1- ”这样的情况，是能够正确计算的，对于“ -1” 这样的情况，就得对横着的线段的右端点+1处理
总结：这类题型非常常见，处理方法也很巧妙，有必要熟练运用。2016百度之星复赛的1003 拍照 也是同类型的题
/**************************************************************
Problem:hdu 5862 Counting Intersections
User: youmi
Language: C++
Result: Accepted
Time:2199MS
Memory:9124K
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#define zeros(a) memset(a,0,sizeof(a))
#define ones(a) memset(a,-1,sizeof(a))
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define sclld(a) scanf("%I64d",&a)
#define pt(a) printf("%d\n",a)
#define ptlld(a) printf("%I64d\n",a)
#define rep(i,from,to) for(int i=from;i<=to;i++)
#define irep(i,to,from) for(int i=to;i>=from;i--)
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define lson (step<<1)
#define rson (lson+1)
#define esp 1e-6
#define oo 0x3fffffff
#define TEST cout<<"*************************"<<endl
const double pi=*atan(1.0); using namespace std;
typedef long long ll;
template <class T> inline void read(T &n)
{
char c; int flag = ;
for (c = getchar(); !(c >= '' && c <= '' || c == '-'); c = getchar()); if (c == '-') flag = -, n = ; else n = c - '';
for (c = getchar(); c >= '' && c <= ''; c = getchar()) n = n * + c - ''; n *= flag;
}
ll Pow(ll base, ll n, ll mo)
{
if (n == ) return ;
if (n == ) return base % mo;
ll tmp = Pow(base, n >> , mo);
tmp = (ll)tmp * tmp % mo;
if (n & ) tmp = (ll)tmp * base % mo;
return tmp;
}
//*************************** int n;
const int maxn=+;
const ll mod=; typedef struct Point
{
int x,y;
int id;
Point(){};
Point(int _x,int _y,int _id)
{
x=_x,y=_y,id=_id;
}
}point;
typedef struct Line
{
point s,e;
int dir;
Line(){};
Line(point _s,point _e,int _dir)
{
s=_s,e=_e,dir=_dir;
}
}line;
line ln[maxn];
point p[maxn<<];
int y[maxn<<];
bool vis[maxn];
bool cmp(point a,point b)
{
return a.x==b.x?ln[a.id].dir<ln[b.id].dir:a.x<b.x;
}
ll c[maxn<<];
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int val)
{
while(x<=*n)
{
c[x]+=val;
x+=lowbit(x);
}
}
ll query(int x)
{
ll res=;
while(x)
{
res+=c[x];
x-=lowbit(x);
}
return res;
} int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int T_T;
scanf("%d",&T_T);
for(int kase=;kase<=T_T;kase++)
{
sc(n);
int x1,y1,x2,y2;
rep(i,,n)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
if(y1>y2||x1>x2)
{
swap(y1,y2);
swap(x1,x2);
}
p[i*-]=Point(x1,y1,i);
p[i*]=Point(x2,y2,i);
y[i*-]=y1,y[i*]=y2;
if(y1==y2)
ln[i]=Line(Point(x1,y1,i),Point(x2,y2,i),),p[i*].x++;//横线的右端点+1
else if(x1==x2)
ln[i]=Line(Point(x1,y1,i),Point(x2,y2,i),);
}
sort(p+,p++*n,cmp);
sort(y+,y++*n);
int k=unique(y+,y++*n)-(y+);
rep(i,,*n)
p[i].y=lower_bound(y+,y++k,p[i].y)-(y);
rep(i,,n)
ln[i].s.y=lower_bound(y+,y++k,ln[i].s.y)-(y),ln[i].e.y=lower_bound(y+,y++k,ln[i].e.y)-(y);
ll ans=;
zeros(c);
zeros(vis);
rep(i,,*n)
{
int dir=ln[p[i].id].dir;
if(dir==)
{
if(vis[p[i].id])
update(p[i].y,-);
else
{
vis[p[i].id]=;
update(p[i].y,);
}
}
else
{
if(vis[p[i].id])
continue;
vis[p[i].id]=;
ll temp=query(ln[p[i].id].e.y);
temp-=query(ln[p[i].id].s.y-);
ans+=temp;
}
}
ptlld(ans);
}
}