poj3304-Segments-判断直线和线段是否相交

最新推荐文章于 2019-07-14 20:40:10 发布

原创最新推荐文章于 2019-07-14 20:40:10 发布 · 379 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#计算几何

计算几何专栏收录该内容

4 篇文章

订阅专栏

本文深入探讨了一种特定算法在处理复杂数据集时的效能与优化策略，结合实际案例分析了该算法在不同场景下的表现，并提出了针对性的改进措施。通过对比分析，揭示了算法在提高数据处理效率、提升准确性方面的潜力与挑战。

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：是否存在一条直线，使得所有线段在其上的投影交于同一点
相当于求是否存在一条直线l交于所有线段，作直线l的垂线m，则m即为所求的直线
将所有线段的两个端点列出来，连接任意两个端点，形成一条直线，然后判断此直线和所有线段是否相交

#include<stdio.h>
#include<math.h>
const double eps=1e-8;
struct point{
    double x,y;
}Seg[1000];
int equ(point a,point b)
{
    double dis;
    dis=(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
    if(sqrt(dis) < eps) return 0;
    else return 1;
}
double fun(point a,point b,point c)
{
    return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
}
int judge(int n)
{
    for(int i=0; i < n ; i++)
    {
        for(int j=i+1 ; j < n ; j++)
        {
            if(equ(Seg[i],Seg[j])==0) continue;
            for(int k=0; k < n ; k+=2)
            {
                double d1,d2;
                d1=fun(Seg[i],Seg[j],Seg[k]);
                d2=fun(Seg[i],Seg[j],Seg[k+1]);
                if(d1*d2>0)
                    break;
                if(k==n-2)
                    return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int T,n;
    double x1,y1,x2,y2;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int j=0;
        for(int i=0 ; i < n ; i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            Seg[j].x=x1; Seg[j].y=y1;
            Seg[j+1].x=x2; Seg[j+1].y=y2;
            j+=2;
        }
        if(judge(2*n)) printf("Yes!\n");
        else printf("No!\n");
    }
    return 0;
}