POJ 1228 Grandpa's Estate -

题目地址：http://poj.org/problem?id=1228

因为每条边至少要有3个点，所以要判断每条边有几个点共线

方法是求出凸包的点，再任意枚举凸包上相邻的两个点，再遍历其他点利用叉积看是否与它共线

要注意：但点数小于6是肯定不可能的，还有所有点都共线的情况也是不可能的

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
#define Vector Point
const double EPS=1e-6;
int Sign(double x){
if(fabs(x)<EPS) return 0;
return x < 0 ? -1 : 1 ;
}
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);
}
bool operator < (const Point& p) const {
if(Sign(x-p.x)==0) return y<p.y;
return x<p.x;
}
};
double cross(const Vector& p1,const Vector& p2){
return p1.x*p2.y-p1.y*p2.x;
}

vector<Point> points;
vector<Point> stack;

bool Graham()
{
if(points.size()<6) return false; //少于6个点是不可能的
stack.clear();

sort(points.begin(),points.end());
stack.push_back(points[0]);
stack.push_back(points[1]);

int n=points.size();
for(int i=2;i<n;i++)
{
while(stack.size()>1){
Point p2=*(stack.end()-1);
Point p1=*(stack.end()-2);
if(Sign(cross(p2-p1,points[i]-p2))<0)
stack.pop_back();
else break;
}
stack.push_back(points[i]);
}

int size=stack.size();
stack.push_back(points[n-2]);
for(int i=n-3;i>=0;i--)
{
while(stack.size()>size){
Point p2=*(stack.end()-1);
Point p1=*(stack.end()-2);
if(Sign(cross(p2-p1,points[i]-p2))<0)
stack.pop_back();
else break;
}
stack.push_back(points[i]);
}
stack.pop_back();
return true;
}
int main()
{
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;
points.clear();
for(int i=0;i<n;i++)
{
int x,y;
cin>>x>>y;
points.push_back(Point(x,y));
}

int ok=0;
if(Graham())
for(int i=0;i<stack.size();i++)
{
Point p1=stack[i],p2=stack[(i+1)%stack.size()];
ok=0;
for(int j=0;j<stack.size();j++)
{
if(j==i||j==((i+1)%stack.size())) continue;
if(Sign(cross(p2-p1,stack[j]-p1))==0) ok++; //若有共线的点就加一
}
if(ok==0||ok==stack.size()-2) break;
}
cout<<((ok==0||ok==stack.size()-2)?"NO":"YES")<<endl;
} //不存在共线点||所有点都共线

return 0;
}