2019东北赛c

思路
这道题思路很简单，也就是直接把所有直线表示成点斜式（如果斜率不存在，另外判断）
但是有一个小技巧不注意的话，就会T
两个分数比较大小的时候，可以先预处理分母，全部变为正数，然后直接分母交叉相乘
$a/b与c/d比较大小的时候，先把分母全部化成正数，然后直接判断ad与bc的大小$

AC代码

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;
struct node{
	bool has_k;
	ll pointx;
	ll yu, yd;
	ll up;
	ll down;
}a[maxn];
ll num[maxn];
///斜率从小到大排序
bool cmp(node b, node c){
	if(!b.has_k && !c.has_k){
		return b.pointx < c.pointx;
	}
	if(!b.has_k && c.has_k) return false;
	if(b.has_k && !c.has_k) return true;
	if(b.up * c.down == c.up * b.down){
		return b.yu * c.yd < c.yu * b.yd;
	}
	return b.up * c.down < c.up * b.down;
}
int main(){
   int t;
   scanf("%d", &t);
   while(t--){
	int n;
	scanf("%d", &n);
	for(int i = 1; i <= n; i++){
		ll x1, y1, x2, y2;
		scanf("%lld%lld%lld%lld", &x1, &y1, &x2, &y2);
		ll u = y2 - y1, d = x2 - x1;///斜率不为0
		if(d){
			if(d < 0){
				a[i].up = -u;
				a[i].down = -d;
			}
			else{
				a[i].up = u;
				a[i].down = d;
			}
			a[i].has_k = true;
			if(y1 == y2){
				a[i].yu = y1;
				a[i].yd = 1;
			}
			else{
				ll u = x2 * y1 - x1 * y2;
				ll d = x2 - x1;
				if(d < 0){
					a[i].yu = -u;
					a[i].yd = -d;
				}
				else{
					a[i].yu = u;
					a[i].yd = d;
				}
			}
		}
		else{
			a[i].has_k = false;
			a[i].pointx = x1;
		}
	}
	sort(a + 1, a + n + 1, cmp);
	int cnt = 0;
	bool has_k = a[1].has_k;
	ll ans = 0, base = 1;
	ll u1, d1;///斜率
	ll u2, d2;///如果斜率存在与y轴交点
	num[++cnt] = 1;
	if(has_k){
		u1 = a[1].up;
		d1 = a[1].down;
		u2 = a[1].yu;
		d2 = a[1].yd;
	}
	else{
		u2 = a[1].pointx;
	}
	ll sum = n;
	for(int i = 2; i <= n; i++){
		if(!has_k || (has_k && a[i].has_k && (u1 * a[i].down == a[i].up * d1))){
			num[cnt]++;
			if(!has_k && a[i].pointx == u2){
				base++;
				if(i == n){
					ans +=  (base - 1) * base / 2;
				}
			}///交点相同
			else if(has_k && a[i].yu * d2 == u2 * a[i].yd){
				//printf("!!!");
				base++;
				if(i == n){
					ans += (base - 1) * base / 2;
				}
			}
			else{
				//printf("!!!\n");
				ans += (base - 1) * base / 2;
				base = 1;
				if(a[i].has_k){
					u2 = a[i].yu;
					d2 = a[i].yd;
				}
				else{
					u2 = a[i].pointx;
				}
			}
		}///斜率相同
		else{
			ans += num[cnt] * (sum - num[cnt]);
			sum -= num[cnt];
			ans += (base - 1) * base / 2;
			base = 1;
			num[++cnt] = 1;
			if(a[i].has_k){
				u1 = a[i].up;
				d1 = a[i].down;
				u2 = a[i].yu;
				d2 = a[i].yd;
			}
			else{
				has_k = false;
				u2 = a[i].pointx;
			}
		}
	}
	printf("%lld\n", ans);
   return 0;
}