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  • poj 3304 判断是否存在一条直线与所有线段相交

    Segments
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 8579   Accepted: 2608

    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!

    题目大意:问是否存在一条直线所有线段在它上面的投影至少有一个公共交点,等同与这条直线的垂线与所有线段都有交点。即求是否有一条与所有线段相交。两两枚举线段四个端点两两成四条直线,
    若所有的线段的两个端点分别在直线的两边(只要不是在同一边就行,在直线上也可以),那么说明存在这么一条直线。
    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <vector>
    #include <algorithm>
    using namespace std;
    
    struct Point{
        double x,y;
        Point(){}
        Point(double x,double y):x(x),y(y){}
    };
    
    struct Segment{
        Point a,b;
    };
    
    typedef Point Vector;
    Vector operator -(const Point &A,const Point &B){ return Vector(A.x-B.x,A.y-B.y);}
    bool operator < (const Point &a,const Point &b)
    {
        return a.x<b.x||(a.x==b.x&&a.y<b.y);
    }
    const double eps=1e-10;
    
    int dcmp(double x)
    {
        if(fabs(x)<eps) return 0;
        else return x<0?-1:1;
    }
    
    bool operator == (const Point &a,const Point &b){
        return (dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0);
    }
    double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}//点积
    double Length(Vector A){return sqrt(Dot(A,A));}//向量长度
    //两向量的夹角
    double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}
    double Cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x;}//叉积
    
    vector<Segment> S;
    
    bool judge(Point a,Point b)
    {
        if(a == b) return false;//a,b属于同一个点,一个点不能确定一条直线
        int i,n=S.size();
        Vector v1=b-a,v2,v3;
        for(i=0;i<n;i++)
        {
            v2=S[i].a-a;
            v3=S[i].b-a;
            if(dcmp(Cross(v1,v2)*Cross(v1,v3)) > 0) return false;
        }
        return true;
    }
    
    bool solve()
    {
        int i,j,n=S.size();
        for(i=0;i<n;i++)
        {
            for(j=i+1;j<n;j++)
            if(judge(S[i].a,S[j].a) || judge(S[i].a,S[j].b) || judge(S[i].b,S[j].a) || judge(S[i].b,S[j].b))
                  return true;
        }
        return false;
    }
    int main()
    {
        int T,n,i;
        Segment s;
        scanf("%d",&T);
        while(T--)
        {
            S.clear();
            scanf("%d",&n);
            for(i=0;i<n;i++)
            {
                scanf("%lf %lf %lf %lf",&s.a.x,&s.a.y,&s.b.x,&s.b.y);
                S.push_back(s);
            }
            if(n==1) printf("Yes!
    ");
            else if(solve()) printf("Yes!
    ");
            else printf("No!
    ");
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/xiong-/p/3421704.html
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