百练 1039 Pipe -

题目地址：http://www.bailian.openjudge.cn/practice/1039

依次枚举上面和下面的任意两点所组成的直线，然后依次判断是否与每个拐点的竖线相交

判断是否相交的方法是算出该点处的y值并比较

#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
const double EPS=1e-6;
#define Vector Point
int n; double mx;

bool IsZero(double x){  
	return -EPS<x&&x<EPS;
}
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);
	} 
	Point operator + (const Vector& v){
		return Point(v.x+x,v.y+y);
	}
	
};
vector<Point> points;

Vector operator * (double f,Vector v){
	return Vector(f*v.x,f*v.y);
}

struct Segment{
	Point x,y;
	double getY(double x1){
		return (x1-x.x)/(y.x-x.x)*(y.y-x.y) + x.y;
	}
	Segment(Point x,Point y):x(x),y(y) {}
};
double cross(Vector a,Vector b){
	return a.x*b.y-a.y*b.x;
}
Point getPoint(Segment a,Segment b){
	Point p1=a.x,p2=a.y,p3=b.x,p4=b.y;
	double x=cross(p3-p1,p2-p1);
	double y=cross(p2-p1,p4-p1);
	if(IsZero(x+y))  return Point(0,0);  //平行
	return p3+x/(x+y)*(p4-p3);
}

bool solved(Point x,Point y){
	for(int k=0;k<n;k++)
	{
		Point p1=points[k],p2=p1; p2.y--;
		double yk=Segment(x,y).getY(p1.x);  //求出在该x点的y值 
		if(yk<p2.y-EPS||yk>p1.y+EPS)
		{
			if(!k) return false;      //若第一个拐点都过不了，直接返回 
			Point p3=points[k-1],p4=p3; p4.y--;
			mx=max(mx,getPoint(Segment(x,y),Segment(p1,p3)).x);
			mx=max(mx,getPoint(Segment(x,y),Segment(p2,p4)).x);
			return false;
		}
		mx=max(mx,p1.x);
	}
	return true;
} 
int main()
{
	while(scanf("%d",&n)&&n)
	{
		double x,y;
		points.clear();
		for(int i=0;i<n;i++){
			scanf("%lf%lf",&x,&y);
			points.push_back(Point(x,y));
		}
		
		int ok=1; mx=points[0].x;
		if(n>2)
		for(int i=0;i<n;i++)  //枚举上面的点和下面的点 
		{
			for(int j=0;j<n;j++)
			{
				if(j==i) continue; 
				Point x=points[i],y=points[j]; y.y--;
				
				ok=0;
				if(solved(x,y)) ok=1;
				if(ok) break;
			}
			if(ok) break;
		}
		if(ok) cout<<"Through all the pipe."<<endl;
		else printf("%.2lf\n",mx);
	}
	return 0;
}