POJ 3304 Segments （直线和线段相交判断）

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7739		Accepted: 2316

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

 

 

题目大意：给出n条线段两个端点的坐标，问所有线段投影到一条直线上，如果这些所有投影至少相交于一点就输出Yes！，否则输出No！。
解题思路：如果有存在这样的直线，过投影相交区域作直线的垂线，该垂线必定与每条线段相交，问题转化为问是否存在一条线和所有线段相交

 

直线肯定经过两个端点。

枚举端点，判断直线和线段是否相交。

 

细节要注意，判断重合点。

 

还有就是加入只有一条线段的话，刚好直线是过同一条直线的。

所以保险的做法是枚举所有的两个端点，包括同一条直线的。

 

/************************************************************

 * Author        : kuangbin

 * Email         : [email protected] 

 * Last modified : 2013-07-13 20:57

 * Filename      : POJ3304Segments.cpp

 * Description   : 

 * *********************************************************/



#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <queue>

#include <map>

#include <vector>

#include <set>

#include <string>

#include <math.h>



using namespace std;



const double eps = 1e-8;

int sgn(double x)

{

    if(fabs(x) < eps)return 0;

    if(x < 0) return -1;

    return 1;

}

struct Point

{

    double x,y;

    Point(){}

    Point(double _x,double _y)

    {

        x = _x;y = _y;

    }

    Point operator -(const Point &b)const

    {

        return Point(x - b.x,y - b.y);

    }

    double operator ^(const Point &b)const

    {

        return x*b.y - y*b.x;

    }

    double operator *(const Point &b)const

    {

        return x*b.x + y*b.y;

    }

};

struct Line

{

    Point s,e;

    Line(){}

    Line(Point _s,Point _e)

    {

        s = _s;e = _e;

    }

};

double xmult(Point p0,Point p1,Point p2) //p0p1 X p0p2

{

    return (p1-p0)^(p2-p0);

}

bool Seg_inter_line(Line l1,Line l2) //判断直线l1和线段l2是否相交

{

    return sgn(xmult(l2.s,l1.s,l1.e))*sgn(xmult(l2.e,l1.s,l1.e)) <= 0;

}

double dist(Point a,Point b)

{

    return sqrt( (b - a)*(b - a) );

}

const int MAXN = 110;

Line line[MAXN];

bool check(Line l1,int n)

{

    if(sgn(dist(l1.s,l1.e)) == 0 )return false;

    for(int i = 0;i < n;i++)

        if(Seg_inter_line(l1,line[i]) == false)

            return false;

    return true;

}

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);

    int n;

    int T;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        double x1,y1,x2,y2;

        for(int i = 0; i < n;i++)

        {

            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);

            line[i] = Line(Point(x1,y1),Point(x2,y2));

        }

        bool flag = false;

        for(int i = 0;i < n;i++)

            for(int j = 0; j < n;j++)

                if(check(Line(line[i].s,line[j].s),n) || check(Line(line[i].s,line[j].e),n)

                        || check(Line(line[i].e,line[j].s),n) || check(Line(line[i].e,line[j].e),n) )

                {

                    flag = true;

                    break;

                }

        if(flag)

            printf("Yes!\n");

        else printf("No!\n");

    }

    return 0;

}