2019东北赛c

C. Line-line Intersection

time limit per test

6.0 s

memory limit per test

512 MB

input

standard input

output

standard output

There are nn lines l1,l2,…,lnl1,l2,…,ln on the 2D-plane.

Staring at these lines, Calabash is wondering how many pairs of (i,j)(i,j) that 1≤i<j≤n1≤i<j≤n and li,ljli,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)T(1≤T≤1000), denoting the number of test cases.

In each test case, there is one integer n(1≤n≤100000)n(1≤n≤100000) in the first line, denoting the number of lines.

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

It is guaranteed that ∑n≤106∑n≤106.

Output

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

Example

input

Copy

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

Copy

1
0
1

当小数被卡精度之时

我们可以考虑用结构体储存一个分数

内非别有 分子分母

在排序时我们只需要按照乘法进行排序即可

但注意的是 在分母为零之时这种情况我们一定要特殊判断

#include<bits/stdc++.h>
using namespace std;
struct fuck
{
    long long x, y, x1, x2, y1, y2;
    long long b;
} node[100005];
int cmp(fuck n1, fuck n2)
{
    if(n1.x * n2.y == n2.x * n1.y)
    {
        return n1.b<n2.b;
    }
    return n1.x * n2.y < n2.x * n1.y;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++)
        {
            scanf("%I64d%I64d%I64d%I64d", &node[i].x1, &node[i].y1, &node[i].x2, &node[i].y2);
            long long x = node[i].x2 - node[i].x1;
            long long y = node[i].y2 - node[i].y1;
            if(x<0)
            {
                x=-x;
                y=-y;
            }
            node[i].x = x;
            node[i].y = y;
            x=abs(x);
            y=abs(y);
            node[i].x/=__gcd(x,y);
            node[i].y/=__gcd(x,y);
            if(node[i].x==0) node[i].b=node[i].x1,node[i].y=1;
            else if(node[i].y==0) node[i].b=node[i].y1,node[i].x=1;
            else node[i].b=node[i].x*node[i].y2-node[i].y*node[i].x2;
        }
        sort(node + 1, node + n + 1, cmp);
        long long ans=0;
        long long cnt=1;
        long long num=0;
        for(int i=1;i<n;i++)
        {
            if(node[i].x==node[i+1].x&&node[i].y==node[i+1].y)
            {
                cnt++;
                if(node[i].b==node[i+1].b)
                {
                    num++;
                    ans+=num;
                }
                else
                {
                    num=0;
                }
            }
            else
            {
                ans+=cnt*(n-i);
                num=0;
                cnt=1;
            }
        }
        printf("%I64d
",ans);
    }
}