POJ3304 思维判断线段和直线相交

Posted 2021-10-23 hesorchen

题目

Segments

给出n个线段，问是否存在一条直线，使得n个线段在这条直线上的投影会交于一点。

解题思路

假设交点为x，过x作这条直线的垂线l，那么必然l经过所有线段。所以问题转化为，是否存在一条直线和n条线段相交。

代码

#include <iostream>
#include <vector>
using namespace std;

const int N = 1e5 + 5;

typedef struct Point
{
    double x, y;
    Point operator-(const Point temp)
    {
        Point T;
        T.x = x - temp.x;
        T.y = y - temp.y;
        return T;
    }
    bool operator==(const Point temp)
    {
        return x == temp.x && y == temp.y;
    }
} Vector;

double cross(Vector A, Vector B)
{
    return A.x * B.y - B.x * A.y;
}

const double eps = 0;

int sign(double x)
{
    if (x < eps)
        return -1;
    else if (x > eps)
        return 1;
    return 0;
}
Point Start[N], End[N];
bool f(Point a1, Point a2, Point b1, Point b2)
{
    double c1 = cross(a2 - a1, b1 - a1), c2 = cross(a2 - a1, b2 - a1);
    return sign(c1) * sign(c2) <= 0;
}

int n;
bool check(Point A, Point B)
{
    if (A == B)
        return 0;
    for (int i = 1; i <= n; i++)
        if (!f(A, B, Start[i], End[i]))
            return 0;
    return 1;
}
void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        scanf("%lf %lf", &Start[i].x, &Start[i].y);
        scanf("%lf %lf", &End[i].x, &End[i].y);
    }
    int flag = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (check(Start[i], Start[j]))
                flag = 1;
            if (check(Start[i], End[j]))
                flag = 1;
            if (check(End[i], End[j]))
                flag = 1;
            if (check(End[i], Start[j]))
                flag = 1;
        }
    }
    puts(flag ? "Yes!" : "No!");
}

int main()
{
    int t;
    cin >> t;
    while (t--)
        solve();
    return 0;
}