zoukankan      html  css  js  c++  java
  • POJ 3304 计算几何

    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!
    ");
          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!
    ");
          else printf("No!
    ");
        }}
        return 0;
    }
    </span>


     

  • 相关阅读:
    hdu 1305 Immediate Decodability
    hdu1247 Hat’s Words
    hdu 1251 统计难题 前缀出现次数
    hdu 3308 线段树,单点更新 求最长连续上升序列长度
    java内部类
    Fckeditor编辑器的安全问题
    Net-NTLMv1的利用
    redis提权
    windows认证过程
    c/c++中system函数在Linux和windows下区别
  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254453.html
Copyright © 2011-2022 走看看