poj1039 Pipe 线段与直线的交点

Pipe

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9393		Accepted: 2849

Description

The GX Light Pipeline Company started to prepare bent pipes for the new transgalactic light pipeline. During the design phase of the new pipe shape the company ran into the problem of determining how far the light can reach inside each component of the pipe.
Note that the material which the pipe is made from is not transparent and not light reflecting.



Each pipe component consists of many straight pipes connected tightly together. For the programming purposes, the company developed the description of each component as a sequence of points [x1; y1], [x2; y2], . . ., [xn; yn], where x1 < x2 < . . . xn . These
are the upper points of the pipe contour. The bottom points of the pipe contour consist of points with y-coordinate decreased by 1. To each upper point [xi; yi] there is a corresponding bottom point [xi; (yi)-1] (see picture above). The company wants to find,
for each pipe component, the point with maximal x-coordinate that the light will reach. The light is emitted by a segment source with endpoints [x1; (y1)-1] and [x1; y1] (endpoints are emitting light too). Assume that the light is not bent at the pipe bent
points and the bent points do not stop the light beam.
Input

The input file contains several blocks each describing one pipe component. Each block starts with the number of bent points 2 <= n <= 20 on separate line. Each of the next n lines contains a pair of real values xi, yi separated by space. The last block is denoted
with n = 0.
Output

The output file contains lines corresponding to blocks in input file. To each block in the input file there is one line in the output file. Each such line contains either a real value, written with precision of two decimal places, or the message Through all
the pipe.. The real value is the desired maximal x-coordinate of the point where the light can reach from the source for corresponding pipe component. If this value equals to xn, then the message Through all the pipe. will appear in the output file.
Sample Input

4
0 1
2 2
4 1
6 4
6
0 1
2 -0.6
5 -4.45
7 -5.57
12 -10.8
17 -16.55
0

Sample Output

4.67
Through all the pipe.

问光从管道中最远能照射到哪个位置(以x轴为基准).光不能透过管道也不能反射。

答：光一定是交于某两个端点照射最远，枚举两个点，然后判断线段（管道的上下口径）与直线（光的照射）是否有交点，有交点则能够照射过去，继续下一个口径，直到从另一端出来或者，没有交点，则光停留在上一个管道壁上。更新最远距离即可。

<pre name="code" class="cpp">#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<functional>
#define eps 1e-9
#include<vector>
#define pi acos(-1)
using namespace std;
const double PI = acos(-1.0);
int dcmp( double x ){ if( fabs(x) < eps ) return 0;else return x < 0?-1:1; }
struct point{
    double x,y;
    point( double x = 0,double y = 0 ):x(x),y(y){}
}node[112];
typedef point Vector;
struct segment
{
    point a,b;
	segment(){}
    segment(point _a,point _b){a=_a,b=_b;}
};
struct circle{
    point c; double r;  circle(){}
    circle(point _c, double _r):c(_c),r(_r) {}
    point  PPP(double a)const{return point(c.x+cos(a)*r,c.y+sin(a)*r);}
};
struct line{
    point p,v; double ang;
    line() {}
    line( const point &_p, const point &_v):p(_p),v(_v){ang = atan2(v.y, v.x);}
    inline bool operator < (const line &L)const{return  ang < L.ang;}
};
point operator + (point a,point b){return point( a.x + b.x,a.y + b.y );}
point operator - (point a,point b){return point( a.x - b.x,a.y - b.y );}
point operator * (point a,double b){return point( a.x*b,a.y*b );}
point operator / (point a,double b){ return point( a.x/b,a.y/b );}
bool  operator <  (const point &a, const point &b ){return  a.x <  b.x || (a.x == b.x && a.y < b.y );}
bool  operator == (const point &a, const point &b ){return (dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0 );}
bool  operator != (const point &a,const point &b ){return a == b?false:true;}

double Dot( point a,point b ){return a.x*b.x + a.y*b.y;} // 点到点的距离  //点积 ab=|a||b|cos@
double Length( point a ){return sqrt( Dot( a,a ) );}    // 向量长度
double Angle( point a,point b ){ return acos( Dot(a,b)/Length(a)/Length(b) );} // 两个向量的角度
double D_T_D(const double deg){ return deg/180.0*PI;}  //弧度转换成角度 

// 向量旋转 rad 度数 rad>0逆时针，rad<0顺时针
point Rotate( point a, double rad ){
     return point( a.x*cos(rad)-a.y*sin(rad),a.x*sin(rad)+a.y*cos(rad) );
}
// 向量的 法线向量 的单位向量 ==向量逆时针旋转90度
point Normal( point a ){
     double L = Length(a); 
	 return point(-a.y/L,a.x/L);
}
// 叉积计算   a*b=|a||b|sin@;
double Cross( point a,point b ){
    return a.x*b.y - a.y*b.x;
}
// 获取 两个向量叉积  判断点pot在直线ab的位置，结果为正在左边，结果为负在右边 
double get_Mix( point a,point b,point pot ){
    a.x = a.x - pot.x; a.y = a.y - pot.y;
    b.x = b.x - pot.x; b.y = b.y - pot.y;
    return Cross( a,b );
}
// 直线相交求交点
point get_line_inter( point p,point v, point q,point w ){
    point u = p - q;
    double t = Cross(w,u)/Cross(v,w);
    return p+v*t;
}
// p点到直线 的距离
double dis_p_line( point p,point a,point b ) {
    point v1 = b-a, v2 = p-a;
    return fabs( Cross(v1,v2)/Length(v1) );
}
//点在直线上的投影
inline point GetLineProjection(const point &p,const point &a,const point &b){
    point v=b-a;
    return a+v*(Dot(v,p-a)/Dot(v,v));
}
// 点到线段的距离
double dis_p_segm( point p,point a,point b ){
    if( a == b )return Length( p-a );
    point v1 = b-a,v2 = p-a,v3 = p-b;
    if( dcmp(Dot(v1,v2)) < 0 )return Length(v2);
    else if( dcmp(Dot( v1,v3)) > 0 )return Length(v3);
    else return fabs(Cross( v1,v2 ))/Length(v1);
}
//海伦公式 三条边
double Heron(double a,double b,double c){
    double p=(a+b+c)/2;
    return sqrt(p*(p-a)*(p-b)*(p-c));
}
// 多边形面积 从p[0] 开始，p
 结束
double ploy_area( vector<point>p,int n ){
    double area = 0;
    n=p.size();
    for( int i = 1; i < n-1; i++ )
       area += Cross( p[i]-p[0],p[i+1]-p[0] );
    return area/2.0;
}

// 线段相交判断 先必须去掉不相交的状态；再判断方向
bool get_set(const point& a1, const point& a2, const point& b1, const point& b2) {
  double c1 = Cross(a2-a1,b1-a1), c2 = Cross(a2-a1,b2-a1),
  c3 = Cross(b2-b1,a1-b1), c4=Cross(b2-b1,a2-b1);
  return dcmp(c1)*dcmp(c2)<eps && dcmp(c3)*dcmp(c4)<-eps;
}

//判断两条线段是否相交，且如何相交
int get_set1(segment l1, segment l2) {
    int d1, d2, d3, d4, d5, d6, d7, d8;
    d1 = dcmp(max(l1.a.x, l1.b.x) - min(l2.a.x, l2.b.x));
    d2 = dcmp(max(l2.a.x, l2.b.x) - min(l1.a.x, l1.b.x));
    d3 = dcmp(max(l1.a.y, l1.b.y) - min(l2.a.y, l2.b.y));
    d4 = dcmp(max(l2.a.y, l2.b.y) - min(l1.a.y, l1.b.y));
    d5 = dcmp(get_Mix(l1.a, l2.b,l2.a));
    d6 = dcmp(get_Mix(l1.b, l2.b,l2.a));
    d7 = dcmp(get_Mix(l2.a, l1.b,l1.a));
    d8 = dcmp(get_Mix(l2.b, l1.b,l1.a));
    if (d1 >= 0 && d2 >= 0 && d3 >= 0 && d4 >= 0) {
        if (!d5 && !d6) return 1; //共线相交
        if (d5 * d6 > 0 || d7 * d8 > 0) return 0; //不相交，与下句代码不能交换顺序
        if (!d5 || !d6 || !d7 || !d8) return 2; //交点为端点
        return 3; //规范相交
    }
    return 0;
}
// 线段 直线 平行判断只需要对应向量平行；
bool get_pall( point a,point b,point c,point d ){
    if( Cross( a-b,c-d ) == 0 )return true;
    return false;
}

// 直线  重合判断 只需要 一条直线的两点都在直线方向
bool get_doub( point a,point b,point c,point d ){
    if( Cross( d-b,a-b ) == 0 && Cross( c-b,a-b ) == 0 )return 1;
    return 0;
}
// 获取 线段 交点；依据 叉积判断
point get_pot( point a,point b,point c,point d ){
    point temp;
    temp.x = ( c.x*Cross(b-a,d-a) - d.x*Cross(b-a,c-a) )/( Cross(b-a,d-a) - Cross(b-a,c-a) );
    temp.y = ( c.y*Cross(b-a,d-a) - d.y*Cross(b-a,c-a) )/( Cross(b-a,d-a) - Cross(b-a,c-a) );
    return temp;
}
//获取直线的交点  同时也可以是线段的交点；
point get_ppp( point a,point b,point c,point d ){
    double a0 = a.y - b.y; double b0 = b.x - a.x; double c0 = a.x*b.y - b.x*a.y;
    double a1 = c.y - d.y; double b1 = d.x - c.x; double c1 = c.x*d.y - d.x*c.y;
    double D =  a0*b1 - a1*b0; point temp;
    temp.x = ( b0*c1 - b1*c0 )/D;
    temp.y = ( a1*c0 - a0*c1 )/D;
    return temp;
}
//点pot 是否 在线段 ab 上 只需 叉积等于0  点积等于0
bool online( point a,point b,point pot ){
    if( Cross( a - pot,b - pot ) == 0 && Dot( a - pot,b - pot ) <= 0 )return 1;
    return 0;
}
//p为点,poly为多边形 点在多边形内判定
int isPointInPolygon(const point& p, const vector<point> poly){
  int n = poly.size();
  int wn = 0;
  for(int i = 0; i < n; i++){
    const point& p1 = poly[i];
    const point& p2 = poly[(i+1)%n];
    if(p1 == p || p2 == p || online(p1, p2, p)) return -1; 
    int k = dcmp(Cross(p2-p1, p-p1));
    int d1 = dcmp(p1.y - p.y);
    int d2 = dcmp(p2.y - p.y);
    if(k > 0 && d1 <= 0 && d2 > 0) wn++;
    if(k < 0 && d2 <= 0 && d1 > 0) wn--;
  }
  if (wn != 0) return 1; //内部 
  return 0; //外部 
}
int top,res[1123456]; // 凸包 ( 起点 0 ) ( n 个点 ) 自己写的，，需要改进 改进；
void GRA( int n )
{
    sort( node,node+n ); // 先排序
    top = 1; res[0] = 0; res[1] = 1;// 从第0位开始放；前两位不管
    for( int i = 2; i <= n; i++ ){
        while( top  && get_Mix( node[i],node[res[top]],node[res[top-1]] ) > 0 )top--;
        res[++top] = i;
    }
    int k = top;
    for( int i = n-2; i >=  0; i-- ){
        while( top > k && get_Mix( node[i],node[res[top]],node[res[top-1]] ) > 0 )top--;
        res[++top] = i;
    }
    top--; // 会添加进去最后一个点
}
//求两圆相交
int C_T_C( circle c1,circle c2,point &p1,point &p2 ){
    double d = Length( c1.c- c2.c );
    if( dcmp( d ) == 0 ) {
        if( dcmp( c1.r-c2.r ) == 0 ) return -1;//两圆重合
        return 0;
    }
    if( dcmp( c1.r + c2.r - d ) < 0 ) return 0;
    if( dcmp( fabs( c1.r - c2.r ) - d ) > 0 ) return 0;
    double a = Angle( c2.c - c1.c,point( 1,0 ) );
    double da = acos(( c1.r * c1.r + d * d - c2.r * c2.r )/( 2 * c1.r * d ) );
    p1 = c1.PPP( a - da ); p2 = c1.PPP( a + da );
    if( p1 == p2 ) return 1;
    return 2;
}
//圆与直线交点 返回交点个数
int C_T_L( line L,circle C,point &p1,point &p2){
    double a = L.v.x, b = L.p.x - C.c.x, c = L.v.y, d = L.p.y-C.c.y;
    double e = a*a + c*c, f = 2*(a*b+c*d), g = b*b + d*d -C.r*C.r;
    double delta = f*f - 4*e*g;
    if( dcmp(delta) < 0 )  return 0;//相离
    if( dcmp(delta) == 0 ) {//相切
        p1 = p1 = L.p + L.v*( -f/(2*e) );
        return 1;
    }//相交
    p1 = ( L.p + L.v * ( -f-sqrt(delta) )/( 2*e ) );
    p2 = ( L.p + L.v * ( -f+sqrt(delta) )/( 2*e ) );
    return 2;
}
//点与圆的切线；
int get_P_C_inter( point p,circle c, point *v )
{
    point u = c.c - p; double dist = Length(u);
    if( dist < c.r )return 0;
    else if( dcmp( dist - c.r) == 0 ){
        v[0] = Rotate( u,PI/2 );
        return 1;
    }else {
        double ang = asin( c.r/dist );
        v[0] = Rotate(u,-ang);
        v[1] = Rotate(u,+ang);
        return 2;
    }
    return -1;
}
//判断点d是否在三角形a,b,c内 
bool inthree(const point a,const point b,const point c,const point d)
{
	point A,B,C;
	A=a-d;
	B=b-d;
	C=c-d;
	double s,s1;
	s=Cross(A,B)/2+Cross(A,C)/2+Cross(B,C)/2;
	s1=Cross(c-b,a-b)/2;
	if (fabs(s-s1)<eps) return true;
	else return false;
}
// 外接圆圆心。假定三点不共线
point get_circumscribed_center(point p1, point p2, point p3) {
  double bx = p2.x - p1.x;
  double by = p2.y - p1.y;
  double cx = p3.x - p1.x;
  double cy = p3.y - p1.y;
  double d = 2 * (bx * cy - by * cx);
  point p;
  p.x = (cy * (bx * bx + by * by) - by * (cx * cx + cy * cy)) / d + p1.x;
  p.y = (bx * (cx * cx + cy * cy) - cx * (bx * bx + by * by)) / d + p1.y;
  return p;
}
double CS(circle a,circle b){    //求两圆相交面积 
	double d=sqrt((a.c.x-b.c.x)*(a.c.x-b.c.x)+(a.c.y-b.c.y)*(a.c.y-b.c.y));
	if (a.r<b.r){
		circle g;
		g=a;
		a=b;
		b=g;
	}
	if (a.r+b.r<=d) return 0; //两圆相离 
	if (a.r-b.r>=d){
		return (pi*b.r*b.r);
	}  //大圆包含小圆 
	double a1=acos((d*d+a.r*a.r-b.r*b.r)/(2*a.r*d));
	double b1=acos((d*d+b.r*b.r-a.r*a.r)/(2*b.r*d));
	double s=a1*a.r*a.r+b1*b.r*b.r;
	double s1=a.r*d*sin(a1);
	return s-s1;
}
//直线与线段是否相交 
bool go(line a,segment b){
	double d1=Cross(a.v-a.p,b.a-a.p);
	double d2=Cross(a.v-a.p,b.b-a.p);
	if (dcmp(d1*d2)<=0) return true;
	return false;
}
point p[25],pp[25];
int vis[50][2];
int main(){
	int n,i,j,k;
	while (~scanf("%d",&n) && n){
		for (i=1;i<=n;i++){
			scanf("%lf%lf",&p[i].x,&p[i].y);
			pp[i].x=p[i].x;
			pp[i].y=p[i].y-1;
		}
		double mx=p[2].x;
		int flag=0;
		for (i=1;i<=n&&!flag;i++){
			for (j=1;j<=n&&!flag;j++)
				if (i!=j){
					for (k=1;k<=n;k++)
						if (!go(line(p[i],pp[j]),segment(p[k],pp[k]))) break;
					if (k==n+1) flag=true;
					else if (k>max(i,j)){
						point gg=get_ppp(p[i],pp[j],p[k-1],p[k]);
						if (mx<gg.x) mx=gg.x;
						gg=get_ppp(p[i],pp[j],pp[k-1],pp[k]);
						if (mx<gg.x) mx=gg.x; 
					}
				} 
		}
		if (flag) cout<<"Through all the pipe."<<endl;
		else printf("%.2f\n",mx);
	}
}