POJ3304(KB13-C 计算几何)

Segments

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15335		Accepted: 4862

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

 

询问是否存在一条直线与所有线段相交

暴力线段的断点构成直线，判断是否可行。

//2017-08-30
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

const int N = 110;
const double EPS = 1e-8;

struct Point{
    double x, y;
    Point(){}
    Point(double _x, double _y):x(_x), y(_y){}
    Point(const Point &p):x(p.x), y(p.y){}
    //a-b为向量ba
    Point operator- (const Point &b) const {
        return Point(x-b.x, y-b.y);
    }
    //向量叉积
    double operator^ (const Point &b) const {
        return x*b.y - y*b.x;
    }
    //向量点积
    double operator* (const Point &b) const {
        return x*b.x + y*b.y;
    }
};

struct Line{
    Point a, b;
    Line(){}
    Line(Point _a, Point _b):a(_a), b(_b){}
    void set_a_b(const Point &_a, const Point &_b){
        a = _a;
        b = _b;
    }
}seg[N];

//三态函数
int sgn(double x){
    if(fabs(x) < EPS)return 0;
    if(x < 0)return -1;
    else return 1;
}

//input：Point a；Point b
//output：distance a、b两点间的距离
//note：该函数取名distance会编译错误
double dist(Point a, Point b){
    return sqrt((a-b)*(a-b));
}

//input：l1 直线l1；l2 线段l2
//output：true 直线l1与线段l2相交；false 直线l1与线段l2不想交
bool seg_inter_line(Line l1, Line l2){
    return sgn((l2.a-l1.b)^(l1.a-l1.b))*sgn((l2.b-l1.b)^(l1.a-l1.b)) <= 0;
}

int n;
//input：直线line
//output：true 直线与所有线段都相交
bool all_inter(Line line){
    for(int i = 0; i < n; i++)
          if(!seg_inter_line(line, seg[i]))
              return false;
    return true;
}

bool check(){
    Line line;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(i == j)continue;
            if(dist(seg[i].a, seg[j].a) > EPS){
                line.set_a_b(seg[i].a, seg[j].a);
                if(all_inter(line))return true;
            }
            if(dist(seg[i].a, seg[j].b) > EPS){
                line.set_a_b(seg[i].a, seg[j].b);
                if(all_inter(line))return true;
            }
            if(dist(seg[i].b, seg[j].a) > EPS){
                line.set_a_b(seg[i].b, seg[j].a);
                if(all_inter(line))return true;
            }
            if(dist(seg[i].b, seg[j].b) > EPS){
                line.set_a_b(seg[i].b, seg[j].b);
                if(all_inter(line))return true;
            }
        }
    }
    return false;
}

int main()
{
    std::ios::sync_with_stdio(false);
    //freopen("inputC.txt", "r", stdin);
    int T;
    cin>>T;
    while(T--){
        cin>>n;
        for(int i = 0; i < n; i++){
            cin>>seg[i].a.x>>seg[i].a.y>>seg[i].b.x>>seg[i].b.y;
        }
        if(check() || n == 1 || n == 2)cout<<"Yes!"<<endl;
        else cout<<"No!"<<endl;
    }

    return 0;
}