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 = ;
 const double EPS = 1e-;

 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 ;
     if(x < )return -;
     else return ;
 }

 //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)) <= ;
 }

 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 = ; i < n; i++)
           if(!seg_inter_line(line, seg[i]))
               return false;
     return true;
 }

 bool check(){
     Line line;
     for(int i = ; i < n; i++){
         for(int j = ; 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 = ; i < n; i++){
             cin>>seg[i].a.x>>seg[i].a.y>>seg[i].b.x>>seg[i].b.y;
         }
         if(check() || n ==  || n == )cout<<"Yes!"<<endl;
         else cout<<"No!"<<endl;
     }

     return ;
 }