[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;
}
posted @ 2012-10-15 21:22  杂鱼  阅读(599)  评论(0编辑  收藏  举报