POJ 3304 Segments (a straight line intersects the line segment is determined)

Topic links: POJ 3304

Problem Description

Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.

Input

Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing a positive integer n ≤ 100 showing the number of segments. After that, n lines containing four real numbers x1 y1 x2 y2 follow, in which (x1, y1) and (x2, y2) are the coordinates of the two endpoints for one of the segments.

Output

For each test case, your program must output "Yes!", if a line with desired property exists and must output "No!" otherwise. You must assume that two floating point numbers a and b are equal if |a - b| < 10-8.

Sample Input

3
2
1.0 2.0 3.0 4.0
4.0 5.0 6.0 7.0
3
0.0 0.0 0.0 1.0
0.0 1.0 0.0 2.0
1.0 1.0 2.0 1.0
3
0.0 0.0 0.0 1.0
0.0 2.0 0.0 3.0
1.0 1.0 2.0 1.0

Sample Output

Yes!
Yes!
No!

Solution

The meaning of problems

Given \ (n-\) line segments, it is determined whether there is a straight line, such that all line segments projected onto this line are common.

answer

Enumeration ToLeftTest

If a straight line to meet the conditions of existence, so that a vertical line must intersect all segments.

If \ (n \ le 2 \) constant presence.

Enumeration of all straight line between any two points of the line segment endpoints configured, it is determined whether all the segments are intersecting.

The method of determining the line segment intersects with the straight line: the line segment is determined whether the two end points on the same side of the line.

Be careful not to enumerate the same two endpoints.

Code

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
typedef double db;
const db eps = 1e-10;  
const db pi = acos(-1.0);  
const ll inf = 0x3f3f3f3f3f3f3f3f;  
const ll maxn = 1e3 + 10;

inline int dcmp(db x) {
    if(fabs(x) < eps) return 0;
    return x > 0? 1: -1;
}

class Point {
public:
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) {}
    void input() {
        scanf("%lf%lf", &x, &y);
    }
    bool operator<(const Point &a) const {
        return (!dcmp(x - a.x))? dcmp(y - a.y) < 0: x < a.x;
    }
    bool operator==(const Point &a) const {
        return dcmp(x - a.x) == 0 && dcmp(y - a.y) == 0;
    }
    db dis2(const Point a) {
        return pow(x - a.x, 2) + pow(y - a.y, 2);
    }
    db dis(const Point a) {
        return sqrt(dis2(a));
    }

    db dis2() {
        return x * x + y * y;
    }
    db dis() {
        return sqrt(dis2());
    }
    Point operator+(const Point a) {
        return Point(x + a.x, y + a.y);
    }
    Point operator-(const Point a) {
        return Point(x - a.x, y - a.y);
    }
    Point operator*(double p) {
        return Point(x * p, y * p);
    }
    Point operator/(double p) {
        return Point(x / p, y / p);
    }
    db dot(const Point a) {
        return x * a.x + y * a.y;
    }
    db cross(const Point a) {
        return x * a.y - y * a.x;
    }
};
typedef Point Vector;

class Line {
public:
    Point s, e;
    Line() {}
    Line(Point s, Point e) : s(s), e(e) {}
    void input() {
        scanf("%lf%lf%lf%lf", &s.x, &s.y, &e.x, &e.y);
    }
    int toLeftTest(Point p) {
        if((e - s).cross(p - s) > 0) return 1;
        else if((e - s).cross(p - s) < 0) return -1;
        return 0;
    }
};

Line line[maxn];
Point point[maxn];

int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; ++i) {
            line[i].input();
            point[i * 2] = line[i].s;
            point[i * 2 + 1] = line[i].e;
        }
        if(n < 3) {
            printf("Yes!\n");
            continue;
        }
        int fg = 0;
        for(int i = 0; i < 2 * n; ++i) {
            for(int j = i + 1; j < 2 * n; ++j) {
                if(point[i] == point[j]) continue;
                Line l = Line(point[i], point[j]);
                int flag = 1;
                for(int k = 0; k < n; ++k) {
                    if(l.toLeftTest(line[k].s) + l.toLeftTest(line[k].e) && l.toLeftTest(line[k].s) == l.toLeftTest(line[k].e)) {
                        flag = 0;
                        break;
                    }
                }
                if(flag == 1) {
                    printf("Yes!\n");
                    fg = 1;
                    break;
                }
            }
            if(fg == 1) {break;}
        }
        if(!fg) {
            printf("No!\n");
        }
    }
    return 0;
}