傳送門：hdu 5862 Counting Intersections
題意：對於平行於坐標軸的n條線段，求兩兩相交的線段對有多少個，包括十，T型
官方題解：由於數據限制,只有豎向與橫向的線段才會產生交點,所以先對橫向線段按x端點排序,每次加入一個線段,將其對應的y坐標位置+1,當出現一個豎向線段時,查詢它的兩個y端點之間的和即為交點個數.
注意點：對x坐標排序是對所有線段端點排序；因為可能出現 “ 1-1 “ 這樣的情況，所以對於橫著的線段，需要進行首尾x坐標處理；我的方法是對於x坐標，先按大小排序，如果大小相同，按先橫後豎的方式排序，那麼對於 “ 1- ”這樣的情況，是能夠正確計算的，對於“ -1” 這樣的情況，就得對橫著的線段的右端點+1處理
總結：這類題型非常常見，處理方法也很巧妙，有必要熟練運用。2016百度之星復賽的1003 拍照 也是同類型的題
/**************************************************************
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=4*atan(1.0);
using namespace std;
typedef long long ll;
template <class T> inline void read(T &n)
{
char c; int flag = 1;
for (c = getchar(); !(c >= '0' && c <= '9' || c == '-'); c = getchar()); if (c == '-') flag = -1, n = 0; else n = c - '0';
for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0'; n *= flag;
}
ll Pow(ll base, ll n, ll mo)
{
if (n == 0) return 1;
if (n == 1) return base % mo;
ll tmp = Pow(base, n >> 1, mo);
tmp = (ll)tmp * tmp % mo;
if (n & 1) tmp = (ll)tmp * base % mo;
return tmp;
}
//***************************
int n;
const int maxn=100000+10;
const ll mod=1000000007;
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<<1];
int y[maxn<<1];
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<<1];
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int val)
{
while(x<=2*n)
{
c[x]+=val;
x+=lowbit(x);
}
}
ll query(int x)
{
ll res=0;
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=1;kase<=T_T;kase++)
{
sc(n);
int x1,y1,x2,y2;
rep(i,1,n)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
if(y1>y2||x1>x2)
{
swap(y1,y2);
swap(x1,x2);
}
p[i*2-1]=Point(x1,y1,i);
p[i*2]=Point(x2,y2,i);
y[i*2-1]=y1,y[i*2]=y2;
if(y1==y2)
ln[i]=Line(Point(x1,y1,i),Point(x2,y2,i),0),p[i*2].x++;//橫線的右端點+1
else if(x1==x2)
ln[i]=Line(Point(x1,y1,i),Point(x2,y2,i),1);
}
sort(p+1,p+1+2*n,cmp);
sort(y+1,y+1+2*n);
int k=unique(y+1,y+1+2*n)-(y+1);
rep(i,1,2*n)
p[i].y=lower_bound(y+1,y+1+k,p[i].y)-(y);
rep(i,1,n)
ln[i].s.y=lower_bound(y+1,y+1+k,ln[i].s.y)-(y),ln[i].e.y=lower_bound(y+1,y+1+k,ln[i].e.y)-(y);
ll ans=0;
zeros(c);
zeros(vis);
rep(i,1,2*n)
{
int dir=ln[p[i].id].dir;
if(dir==0)
{
if(vis[p[i].id])
update(p[i].y,-1);
else
{
vis[p[i].id]=1;
update(p[i].y,1);
}
}
else
{
if(vis[p[i].id])
continue;
vis[p[i].id]=1;
ll temp=query(ln[p[i].id].e.y);
temp-=query(ln[p[i].id].s.y-1);
ans+=temp;
}
}
ptlld(ans);
}
}
