[报告]ACM-ICPC 2012 Regionals Asia - Changchun I Polaris of Pandora / ZJU 3663

Abstract

ACM-ICPC 2012 Regionals Asia - Changchun I Polaris of Pandora / ZJU 3663

三维计算几何

Body

Source

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3663

Description

一个半径为R的球，给定起点终点的经纬度和飞行高度H。平行光从正北方向往正南方向照射，问飞行路程中被光找到的比例。

Solution

本场悲剧的我们大概也就这水平了……

本质上就是求飞行路径所在的大圆和半径为R的圆柱的交点。建立直角坐标系硬算就好了……

本题精度坑爹（h和R可能相差一万倍），不可以把高度和半径归一化（注意我中间注释的那两行），但是不归一化我算出来的点甚至和起点终点不共面……当然也可能是我姿势不优美……总之给1y的hl大神跪了。

Code

#include <iomanip>

#include <cassert>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <iostream>

#include <algorithm>

using namespace std;



const double EPS = 1e-8;

const double PI = acos(-1.0);



inline int sgn(double d) {

    if (fabs(d)<EPS) return 0;

    return d>0?1:-1;

}



struct sp {

    double x, y, z;

    sp() {}

    sp(double a, double b, double c): x(a), y(b), z(c) {}

    void write() {

        printf("%f %f %f\n", x, y, z);

    }

};



sp operator-(const sp &u, const sp &v) {

    return sp(u.x-v.x, u.y-v.y, u.z-v.z);

}



double dot(const sp &u, const sp &v) {

    return u.x*v.x+u.y*v.y+u.z*v.z;

}



sp det(const sp &u, const sp &v) {

    return sp(u.y*v.z-v.y*u.z,

            u.z*v.x-u.x*v.z,

            u.x*v.y-u.y*v.x);

}



double nrm(const sp &u) {

    return sqrt(u.x*u.x+u.y*u.y+u.z*u.z);

}



double dis(const sp &u, const sp &v) {

    double dx = u.x-v.x;

    double dy = u.y-v.y;

    double dz = u.z-v.z;

    return sqrt(dx*dx+dy*dy+dz*dz);

}



double ang(const sp &u, const sp &v) {

    double d = dot(u, v);

    return acos(d/(nrm(u)*nrm(v)));

}



double r, h, as, ns, at, nt, R;

sp s, t, hmr, mdk;

double a, b, c, tmp;



bool quad(double a, double b, double c, double &x1, double &x2) {

    double delta = b*b-4*a*c;

    if (sgn(delta)<=0) return 0;

    x1 = (-b+sqrt(delta))/(2*a);

    x2 = (-b-sqrt(delta))/(2*a);

    return 1;

}



bool coplane(const sp &a, const sp &b, const sp &c, const sp &d) {

    return sgn(dot(det(c-a, b-a), d-a))==0;

}



bool inside(const sp &s, const sp &p, const sp &t) {

    return sgn(dot(det(s,p),det(s,t)))>=0 && sgn(dot(det(p,t),det(s,t)))>=0;

}



int main() {

    while (cin>>r>>h>>as>>ns>>at>>nt) {

        if (sgn(as)==0 && sgn(at)==0) {

            puts("100.000");

            continue;

        }

        as += PI/2; at += PI/2;

        ns += PI; nt += PI;

        //h /= r; r = 1;

        R = r+h;

        s = sp(R*sin(as)*cos(ns), R*sin(as)*sin(ns), R*cos(as));

        t = sp(R*sin(at)*cos(nt), R*sin(at)*sin(nt), R*cos(at));

        a = s.y*t.z-s.z*t.y; b = s.z*t.x-s.x*t.z; c = s.x*t.y-s.y*t.x;

        tmp = -sqrt(R*R-r*r);

        hmr.z = mdk.z = tmp;

        assert(sgn(a)!=0 || sgn(b)!=0);

        if (sgn(b)) {

            if (quad(a*a+b*b, 2*a*c*tmp, c*c*(R*R-r*r)-b*b*r*r, hmr.x, mdk.x)) {

                hmr.y = (-c*tmp-a*hmr.x)/b;

                mdk.y = (-c*tmp-a*mdk.x)/b;

            }

            else {

                puts("100.000");

                continue;

            }

        }

        else {

            if (quad(a*a+b*b, 2*b*c*tmp, c*c*(R*R-r*r)-a*a*r*r, hmr.y, mdk.y)) {

                hmr.x = (-c*tmp-b*hmr.y)/a;

                mdk.x = (-c*tmp-b*mdk.y)/a;

            }

            else {

                puts("100.000");

                continue;

            }

        }

        //assert(coplane(s, t, hmr, mdk));

        int cnt = 0;

        if (inside(s, hmr, t)) ++cnt;

        if (inside(s, mdk, t)) ++cnt;

        double tot=ang(s, t), dark;

        if (cnt==0) {

            if (inside(hmr, s, mdk) && inside(hmr, t, mdk)) dark = tot;

            else dark = 0;

        }

        else if (cnt==1) {

            if (inside(s, mdk, t)) swap(hmr, mdk);

            assert(inside(s, hmr, t)&&!inside(s, mdk, t));

            if (inside(hmr, s, mdk)) dark = ang(hmr, s);

            else dark = ang(hmr, t);

        }

        else dark = ang(hmr, mdk);

        double ans = (1-dark/tot)*100;

        printf("%.3f\n", ans);

        //cout<<fixed<<setprecision(3)<<ans<<endl;

    }

    return 0;

}