There are n lines l1,l2,…,ln

on the 2D-plane.

Staring at these lines, Calabash is wondering how many pairs of (i,j)

that 1≤i<j≤n and li,lj

share at least one common point. Note that two overlapping lines also share common points.

Please write a program to solve Calabash's problem.

Input

The first line of the input contains an integer T(1≤T≤1000)

, denoting the number of test cases.

In each test case, there is one integer n(1≤n≤100000)

in the first line, denoting the number of lines.

For the next n

lines, each line contains four integers xai,yai,xbi,ybi(|xai|,|yai|,|xbi|,|ybi|≤109). It means li passes both (xai,yai) and (xbi,ybi). (xai,yai) will never be coincided with (xbi,ybi)

.

It is guaranteed that ∑n≤106

.

Output

For each test case, print a single line containing an integer, denoting the answer.

Example Input 3 2 0 0 1 1 0 1 1 0 2 0 0 0 1 1 0 1 1 2 0 0 1 1 0 0 1 1 Output 1 0 1 題解：兩條直線不平行必相交，若平行：若重合答案加1，否則不算。 我們用兩個map來刻畫直線的特性，mp1刻畫a*x+b*y的直線系有多少個，mp2刻畫a*x+b*y=c這一條直線有多少個。 假設當前直線與之前的線段都相交，那么我們需要減去與這條直線平行而不重合的直線。即ans+=i-1+mp2[]-mp1[]. #include<iostream> #include<cstring> #include<string> #include<queue> #include<stack> #include<algorithm> #include<stdio.h> #include<map> #include<set> using namespace std; typedef long long ll; typedef pair<ll,ll>P; typedef pair<pair<ll,ll>,ll>Pi; const int maxn=100010; map<pair<ll,ll>,ll>mp1;//兩個參數a,b，代表形如a*x+b*y=c（c任意）的直線有多少個 map<pair<pair<ll,ll>,ll>,ll>mp2;//三個參數a,b,c,代表形如a*x+b*y=c的直線有多少個，即相同直線有多少個 ll cnt,ans,n; int main() {ios::sync_with_stdio(0);int T;cin>>T;while(T--){ans=cnt=0;cin>>n;mp1.clear(),mp2.clear();for(int i=1;i<=n;i++){ll x1,x2,y1,y2;cin>>x1>>y1>>x2>>y2;ll a=x1-x2,b=y1-y2,c=x1*y2-x2*y1;ll g=__gcd(a,b);a/=g;b/=g;c/=g;mp1[P(a,b)]++;mp2[Pi(P(a,b),c)]++;ans+=i-1+mp2[Pi(P(a,b),c)]-mp1[P(a,b)];//假設與前i-1條邊都相交，需要減去與他平行而不重合的線段 }cout<<ans<<endl;}return 0; }

?

轉載于:https://www.cnblogs.com/cherish-lin/p/11066282.html

總結

以上是生活随笔為你收集整理的C - Line-line Intersection Gym - 102220C（线段相交）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。