poj 1263 Reflections (Simple Geometry)

1263 -- Reflections

　　简单计算几何。题目给出射线以及若干个不相交的圆，求出射线会在哪些圆上反弹，依次写出反弹球的编号。

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>

using namespace std;

const double EPS = 1e-10;
template<class T> T sqr(T x) { return x * x;}
inline int sgn(double x) { return (x > EPS) - (x < -EPS);}
struct Point {
    double x, y;
    Point() {}
    Point(double x, double y) : x(x), y(y) {}
    bool operator < (Point a) const { return sgn(x - a.x) < 0 || sgn(x - a.x) == 0 && y < a.y;}
    bool operator == (Point a) const { return sgn(x - a.x) == 0 && sgn(y - a.y) == 0;}
    Point operator + (Point a) { return Point(x + a.x, y + a.y);}
    Point operator - (Point a) { return Point(x - a.x, y - a.y);}
    Point operator * (double p) { return Point(x * p, y * p);}
    Point operator / (double p) { return Point(x / p, y / p);}
} ;
typedef Point Vec;

inline double cross(Vec a, Vec b) { return a.x * b.y - a.y * b.x;}
inline double cross(Point o, Point a, Point b) { return cross(a - o, b - o);}
inline double dot(Vec a, Vec b) { return a.x * b.x + a.y * b.y;}
inline double dot(Point o, Point a, Point b) { return dot(a - o, b - o);}
inline double veclen(Vec x) { return sqrt(dot(x, x));}
inline Vec normal(Vec x) { return Vec(-x.y, x.x) / veclen(x);}

struct Line {
    Point s, t;
    Line() {}
    Line(Point s, Point t) : s(s), t(t) {}
    Vec vec() { return t - s;}
    Point pt(double x) { return s + vec() * x;}
    Line move(double x) {
        Vec nor = normal(vec());
        return Line(s + nor * x, t + nor * x);
    }
} ;

inline Point llint(Point P, Vec v, Point Q, Vec w) { return P + v * (cross(w, P - Q) / cross(v, w));}
inline Point llint(Line a, Line b) { return llint(a.s, a.vec(), b.s, b.vec());}

struct Circle {
    Point c;
    double r;
    Circle() {}
    Circle(Point c, double r) : c(c), r(r) {}
} ;

void lcint(Line L, Circle C, vector<Point> &sol) {
    Point ip = llint(L, Line(C.c, C.c + normal(L.vec())));
    double dis = veclen(ip - C.c);
    if (sgn(dis - C.r) >= 0) return ;
    Vec u = L.vec() / veclen(L.vec());
    double d = sqrt(sqr(C.r) - sqr(dis));
    sol.push_back(ip + u * d);
    sol.push_back(ip - u * d);
}

Point reflect(Point x, Line L) {
    Vec nor = normal(L.vec());
    Point ip = llint(L, Line(x, x + nor));
    return ip + ip - x;
}

vector<Circle> rec;
Point src;
Vec dir;

const double FINF = 1e100;
inline bool onCircle(Point p, Circle c) { return sgn(veclen(p - c.c) - c.r) == 0;}

void work() {
    int cnt = 0, sz = rec.size();
    vector<Point> tmp;
    Point ip;
    while (true) {
        tmp.clear();
        double d = FINF;
        for (int i = 0; i < sz; i++) lcint(Line(src, src + dir), rec[i], tmp);
        for (int i = 0, sz = tmp.size(); i < sz; i++) {
            double t = (tmp[i].x - src.x) / dir.x;
            if (t > EPS) d = min(d, t);
        }
        if (sgn(d - FINF) >= 0) break;
        cnt++;
        if (cnt > 10) break;
        ip = src + dir * d;
        int mk = -1;
        for (int i = 0; i < sz; i++) if (onCircle(ip, rec[i])) { mk = i; break;}
        if (mk == -1) { puts("shit!!"); while (1) ;}
        printf("%d ", mk + 1);
        dir = reflect(src, Line(ip, rec[mk].c)) - ip;
        src = ip;
    }
    if (cnt > 10) puts("...");
    else puts("inf");
}

int main() {
//    freopen("in", "r", stdin);
    int cas = 1, n;
    double x, y, r;
    while (cin >> n && n) {
        rec.clear();
        while (n--) {
            cin >> x >> y >> r;
            rec.push_back(Circle(Point(x, y), r));
        }
        cin >> src.x >> src.y;
        cin >> dir.x >> dir.y;
        dir = dir / veclen(dir);
        printf("Scene %d
", cas++);
        work();
        puts("");
    }
    return 0;
}

　　速度好慢，整整写了一个小时。对几何模板还是不算非常熟悉，虽然已经能够灵活写出部分基础函数了，但是速度还真是一个大问题。最后结果，因为手多血多个换行PE了一次，然后就AC了~

——written by Lyon