The 13th Chinese Northeast Contest C. Line-line Intersection(平面几何)
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题目描述
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
题目大意
有n条线,给你n条线上的一对点。问这些线之间有多少对交点(一对重合的线算1)。
题目分析
这道题有三种情况:
- 两条直线的斜率k不相同,则这两条直线一定会有一个交点。
- 两条直线的斜率k相同,但是截距b不相同,则两条线一定没有交点。
- 两条直线的斜率k和截距b都相同,则这两条直线有交点。
每读入一条直线,我们只需要算出这一条直线的斜率k和截距b,numk //表示之前读入的直线中斜率也为k的直线个数
| t //表示之前读入的直线中斜率为k并且截距为b的直线个数
那么之前读入的直线中与该条直线相交的直线数为: n u m ( 直 线 总 数 ) − n u m k + t 。 num(直线总数)-numk+t。 num(直线总数)−numk+t。
注:这道题的斜率k和截距b不能用浮点数来存(因为有浮点误差,浮点数不靠谱)
我们可以用pair来存储k:
y
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c
d
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x
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,
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g
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{y/gcd(x,y) , x/gcd(x,y)}
y/gcd(x,y),x/gcd(x,y) (x=x2-x1,y=y2-y1)
用
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(x1*y2-x2*y1)/gcd(x,y)
(x1∗y2−x2∗y1)/gcd(x,y) 来表示b。
代码如下
#include <iostream>
#include <cmath>
#include <cstdio>
#include <set>
#include <string>
#include <cstring>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <bitset>
#define LL long long
#define ULL unsigned long long
#define PII pair<LL,LL>
#define PDD pair<double,double>
#define x first
#define y second
using namespace std;
const int N=1e5+5,INF=1e9+7;
struct Node {
PII k; //表示斜率 k
LL b; //表示截距 b
bool operator< (const Node a)const
{
if(k==a.k) return b<a.b;
return k<a.k;
}
};
int main()
{
cin.tie(0);
ios::sync_with_stdio(false);
int t;
cin>>t;
while(t--)
{
int n;
cin>>n;
map<PII,int> mp; //这个map中只存储斜率,用于查找斜率相同的直线的个数
map<Node,int> s; //这个map中存斜率和截距,用于在查找斜率和截距都相同的线的个数
LL ans=0;
for(int i=1;i<=n;i++)
{
LL x1,x2,y1,y2;
cin>>x1>>y1>>x2>>y2;
LL x=x2-x1,y=y2-y1,c=x1*y2-x2*y1;
LL g=__gcd(x,y);
PII k={x/g,y/g}; c/=g;
int numk=mp[k],t=s[{k,c}]; //算出 numk 和 t
ans+=i-1-numk+t; //记录答案
mp[k]++; //将这条线放入集合中
s[{k,c}]++;
}
cout<<ans<<endl;
}
return 0;
}
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