POJ 3304 计算几何

最新推荐文章于 2022-06-18 12:22:14 发布

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最新推荐文章于 2022-06-18 12:22:14 发布

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原文链接：http://www.cnblogs.com/codeyuan/p/4254453.html

本文提供了一个解决几何问题的算法，旨在判断给定的二维空间内的多个线段是否至少存在一条直线使得所有线段投影后至少有一个公共点。通过输入测试案例的数量、线段数量及线段的端点坐标，输出结果为存在这样的直线则显示Yes!，否则显示No!。该算法采用计算几何方法，包括点与线段之间的相对位置判断和直线的构建。

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Segments
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9564 Accepted: 2943

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!
Source

Amirkabir University of Technology Local Contest 2006

<span style="color:#6600cc;">/**************************************
     author    : Grant Yuan
     time      : 2014/8/19 16:53
     algorithm : 计算几何
     source    : POJ 3304
***************************************/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define MAX 107
#define eps 1e-8
#include<cmath>

using namespace std;

struct Point
{
    double x,y;
    Point(){}
    Point(double _x,double _y)
    {
        x=_x;
        y=_y;
    }
    Point operator -(const Point&b)
    {
        return Point(x-b.x,y-b.y);
    }
    double operator *(const Point&b)
    {
       return x*b.x+y*b.y;
    }
    double operator ^(const Point&b)
    {
        return x*b.y-y*b.x;
    }
    bool operator ==(const Point&b)
    {
        return(x==b.x&&y==b.y);
    }
};

struct Line
{
    Point p1,p2;
    Line(){}
    Line(Point a,Point b){p1=a;p2=b;}
};

Point P[2*MAX];
Line L[MAX];

bool check(Point q0,Point q1,Point q2,Point q3)
{
    if(fabs(q0.x-q1.x)<eps&&fabs(q0.y-q1.y)<eps)
        return false;
    int t1=0,t2=0;
    if(((q1-q0)^(q2-q0))<-eps) t1=-1;
    else if(((q1-q0)^(q2-q0))>eps) t1=1;

     if(((q1-q0)^(q3-q0))<-eps) t2=-1;
    else if(((q1-q0)^(q3-q0))>eps) t2=1;
    if(t1*t2==-1||t1==0||t2==0)
        return true;
    return false;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
      int n;
      scanf("%d",&n);
      for(int i=0;i<2*n;i++)
      {
          scanf("%lf%lf",&P[i].x,&P[i].y);
      }
      if(n==1||n==2) printf("Yes!\n");
      else{
      for(int i=0;i<2*n;i++)
      {
          if(i%2==0) L[i/2].p1=P[i];
          else L[i/2].p2=P[i];
      }
      Point q0,q1;
      bool ans=0;
      for(int i=0;i<2*n-1;i++)
      {
        q0=P[i];
        for(int j=i+1;j<2*n;j++)
        {
        if(P[j].x==q0.x&&P[j].y==q0.y) continue;
        q1=P[j];
        bool flag=1;
        for(int k=0;k<n;k++)
        {
            if(!check(q0,q1,L[k].p1,L[k].p2)){flag=0; break;}
        }
        if(flag) {ans=1;break;}
        }
        if(ans) break;
      }
       if(ans) printf("Yes!\n");
      else printf("No!\n");
    }}
    return 0;
}
</span>

转载于:https://www.cnblogs.com/codeyuan/p/4254453.html