POJ 3304 Segments -

题目地址：http://poj.org/problem?id=3304

任意枚举两点组成的线段是否能与所有线段相交

判断相交的办法是叉积小于等于零，但在叉积之前要排除掉两个一样的点，因为零向量与任何向量叉积都为0

#include<iostream>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
const double EPS=1e-8;
bool IsZero(double x){ //判断x是否为零
return -EPS<x&&x<EPS;
}
#define Vector Point
struct Point{
double x,y;
Point(double x,double y):x(x),y(y){};
Vector operator - (const Point& p){
return Vector(x-p.x,y-p.y);
}
};
vector<Point> points;

double cross(Vector s1,Vector s2){
return s1.x*s2.y-s1.y*s2.x;
}
double dist(Point p1,Point p2){
return hypot(fabs(p1.x-p2.x),fabs(p1.y-p2.y));
}
bool solved(int i,int j){
Point p1=points[i],p2=points[j];
if(IsZero(dist(p1,p2))) return false; //一定要写，不写会错 ，因为一个点与任何线叉乘都为0
for(int k=0;k<points.size();k++){
if(k==i||k==j||(k%2==0?k+1:k-1)==i||(k%2==0?k+1:k-1)==j) continue;
Point p3=points[k],p4=points[(k%2==0?k+1:k-1)];
if(cross(p2-p1,p3-p1)*cross(p2-p1,p4-p1) > EPS) return false;
}
return true;
}

int main()
{
int T;
cin>>T;
while(T--){
int n;
double x1,y1,x2,y2;
points.clear();
cin>>n;
while(n--) {
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
points.push_back(Point(x1,y1));
points.push_back(Point(x2,y2));
}

int ok=1;
if(n!=1)
for(int i=0;i<points.size()-1;i++)
{
for(int j=i+1;j<points.size();j++)
{
ok=0;
if(solved(i,j)) ok=1;
if(ok) break;
}
if(ok) break;
}
cout<<(ok?"Yes!":"No!")<<endl;
}
return 0;
}