Segments - POJ 3304 线段相交

最新推荐文章于 2020-07-19 22:59:12 发布

提比-我有特殊的AC技巧

最新推荐文章于 2020-07-19 22:59:12 发布

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分类专栏： POJ 几何

本文链接：https://blog.youkuaiyun.com/u014733623/article/details/46920211

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本文探讨了一个几何问题，即在二维空间中，通过给定的线段集合，判断是否存在一条直线使得所有线段在该直线上投影后至少有一个公共点。提出了一种算法来解决此问题，并提供了详细的输入输出样例。

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Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10570		Accepted: 3281

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：给出n条线段，判断是否存在有一条直线，满足所有的线段在直线上投影后至少有一个公共点

思路：原命题等价为存在一条直线穿过所有的线段（易知过公共点且垂直于所求直线的直线符合条件，设为直线a），该命题又等价于从所有线段中任选两端点形成的直线存在可以穿过所有的线段的直线（可将a平移至一条线段端点，然后绕这点旋转，使a过另一条线段端点）

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y){}
}p[210];
typedef Point Vector;
double eps=1e-8;
Vector operator + (Vector A,Vector B){return Point(A.x+B.x,A.y+B.y);}
Vector operator - (Vector A,Vector B){return Point(A.x-B.x,A.y-B.y);}
int dcmp(double x){return (x>eps)-(x<-eps);}
double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
int SegmentInterSection(Point a1,Point a2,Point b1,Point b2)
{
    int d1,d2;
    d1=dcmp(Cross(a2-a1,b1-a1));
    d2=dcmp(Cross(a2-a1,b2-a1));
    if((d1^d2)==-2)
      return 1;
    if(d1==0 || d2==0)
      return 2;
    return 0;
}
int T,t,n,m;
bool flag;
bool test(Point a1,Point a2)
{
    int i,j,k;
    if(dcmp(a1.x-a2.x)==0 && dcmp(a1.y-a2.y)==0)
      return 0;
    for(i=1;i<=2*n;i+=2)
       if(SegmentInterSection(a1,a2,p[i],p[i+1])==0)
         return 0;
    return 1;
}
void solve()
{
    int i,j;
    for(i=1;i<=n*2;i++)
       for(j=i+1;j<=n*2;j++)
          if(test(p[i],p[j]))
          {
              flag=1;
              return;
          }
}
int main()
{
    int i,j,k;
    scanf("%d",&T);
    for(t=1;t<=T;t++)
    {
        scanf("%d",&n);
        for(i=1;i<=n*2;i++)
           scanf("%lf%lf",&p[i].x,&p[i].y);
        flag=0;
        solve();
        if(flag)
          printf("Yes!\n");
        else
          printf("No!\n");

    }
}