POJ-3304Segments[计算几何]

Segments

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!

题意：

求n条线段在一条直线上的投影是否有交点

思路：

假设存在这样的一个交点，那么过交点做投影直线的垂线，必定和所有的线段相交，然后就将问题化为构造一条和所有线段相交的直线。通过每次枚举两个线段的

端点作出直线。

#include "stdio.h"
#include "iostream"
#include "algorithm"
#include "cmath"
#include "string.h"
using namespace std;
const int maxn = 100 + 10;
const double eps = 1e-8;
struct Point {
    double x, y;
    Point(double x=0.0, double y=0.0) :x(x), y(y) {}
};
typedef Point Vector;
Vector operator - (Point A,Point B) {return Vector(A.x-B.x,A.y-B.y);}
int dcmp(double x) {if (fabs(x) < eps) return 0;return x < 0? -1: 1;} 
double Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}
double Length(Point A) {return sqrt(Dot(A,A));}
double Cross(Vector A,Vector B) {return A.x*B.y-A.y*B.x;}

int N;
Point p[maxn][4];
bool getcross(Point L, Point S) {
    if (dcmp(Length(S - L)) == 0) return false;
    for (int i = 0; i < N; i++) {
        if (dcmp(Cross(S-L, p[i][0]-L)*Cross(S-L, p[i][1]-L))>0) return false;

    }
    return true;
}
void solve() {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (getcross(p[j][0],p[i][0])||getcross(p[j][1],p[i][0])||
                getcross(p[j][1],p[i][1])||getcross(p[j][0],p[i][1])) {
                printf("Yes!
"); return ;
            }
        }
    }
    printf("No!
");
}
int main(int argc, char const *argv[])
{
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d", &N);
        for (int i = 0; i < N; i++) {
            scanf("%lf%lf%lf%lf", &p[i][0].x, &p[i][0].y, &p[i][1].x, &p[i][1].y);
        }
        solve();
    }
    return 0;
}
/*
1
6
75 93 42 45
47 44 87 30
10 55 36 14
32 81 73 57
29 84 75 23
18 38 99 95
*/