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题意:一根管道,有光源从入口发射,问光源最远到达的地方。
分析:黑书上的例题,解法是枚举任意的一个上顶点和一个下顶点(优化后),组成直线,如果直线与所有竖直线段有交点,则表示能穿过管道。
/************************************************ * Author :Running_Time * Created Time :2015/10/31 星期六 10:28:12 * File Name :POJ_1039.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int N = 1e5 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = acos (-1.0); int dcmp(double x) { //三态函数,减少精度问题 if (fabs (x) < EPS) return 0; else return x < 0 ? -1 : 1; } struct Point { //点的定义 double x, y; Point () {} Point (double x, double y) : x (x), y (y) {} Point operator + (const Point &r) const { //向量加法 return Point (x + r.x, y + r.y); } Point operator - (const Point &r) const { //向量减法 return Point (x - r.x, y - r.y); } Point operator * (double p) const { //向量乘以标量 return Point (x * p, y * p); } Point operator / (double p) const { //向量除以标量 return Point (x / p, y / p); } bool operator < (const Point &r) const { //点的坐标排序 return x < r.x || (x == r.x && y < r.y); } bool operator == (const Point &r) const { //判断同一个点 return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0; } }; typedef Point Vector; //向量的定义 Point read_point(void) { //点的读入 double x, y; scanf ("%lf%lf", &x, &y); return Point (x, y); } double dot(Vector A, Vector B) { //向量点积 return A.x * B.x + A.y * B.y; } double cross(Vector A, Vector B) { //向量叉积 return A.x * B.y - A.y * B.x; } double polar_angle(Vector A) { //向量极角 return atan2 (A.y, A.x); } double length(Vector A) { //向量长度,点积 return sqrt (dot (A, A)); } double angle(Vector A, Vector B) { //向量转角,逆时针,点积 return acos (dot (A, B) / length (A) / length (B)); } Vector rotate(Vector A, double rad) { //向量旋转,逆时针 return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad)); } Vector nomal(Vector A) { //向量的单位法向量 double len = length (A); return Vector (-A.y / len, A.x / len); } Point line_line_inter(Point p, Vector V, Point q, Vector W) { //两直线交点,参数方程 Vector U = p - q; double t = cross (W, U) / cross (V, W); return p + V * t; } double point_to_line(Point p, Point a, Point b) { //点到直线的距离,两点式 Vector V1 = b - a, V2 = p - a; return fabs (cross (V1, V2)) / length (V1); } double point_to_seg(Point p, Point a, Point b) { //点到线段的距离,两点式 if (a == b) return length (p - a); Vector 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); } Point point_line_proj(Point p, Point a, Point b) { //点在直线上的投影,两点式 Vector V = b - a; return a + V * (dot (V, p - a) / dot (V, V)); } bool can_seg_seg_inter(Point a1, Point a2, Point b1, Point b2) { //判断线段相交,两点式 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) < 0 && dcmp (c3) * dcmp (c4) < 0; } bool can_line_seg_inter(Point a1, Point a2, Point b1, Point b2) { double c1 = cross (a2 - a1, b1 - a1), c2 = cross (a2 - a1, b2 - a1); return dcmp (c1 * c2) <= 0; } bool on_seg(Point p, Point a1, Point a2) { //判断点在线段上,两点式 return dcmp (cross (a1 - p, a2 - p)) == 0 && dcmp (dot (a1 - p, a2 - p)) < 0; } double area_triangle(Point a, Point b, Point c) { //三角形面积,叉积 return fabs (cross (b - a, c - a)) / 2.0; } double area_poly(Point *p, int n) { //多边形面积,叉积 double ret = 0; for (int i=1; i<n-1; ++i) { ret += fabs (cross (p[i] - p[0], p[i+1] - p[0])); } return ret / 2; } /* 点集凸包,输入点的集合,返回凸包点的集合。 如果不希望在凸包的边上有输入点,把两个 <= 改成 < */ vector<Point> convex_hull(vector<Point> ps) { sort (ps.begin (), ps.end ()); //x - y排序 ps.erase (unique (ps.begin (), ps.end ()), ps.end ()); //删除重复点 int n = ps.size (), k = 0; vector<Point> qs (n * 2); for (int i=0; i<n; ++i) { while (k > 1 && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-1]) <= 0) k--; qs[k++] = ps[i]; } for (int i=n-2, t=k; i>=0; --i) { while (k > t && cross (qs[k-1] - qs[k-2], ps[i] - qs[k-1]) <= 0) k--; qs[k++] = ps[i]; } qs.resize (k-1); return qs; } struct Circle { Point c; double r; Circle () {} Circle (Point c, double r) : c (c), r (r) {} Point point(double a) { return Point (c.x + cos (a) * r, c.y + sin (a) * r); } }; struct Line { Point p; Vector v; double r; Line () {} Line (const Point &p, const Vector &v) : p (p), v (v) { r = polar_angle (v); } Point point(double a) { return p + v * a; } }; /* 直线相交求交点,返回交点个数,交点保存在P中 */ int line_cir_inter(Line L, Circle C, double &t1, double &t2, vector<Point> &P) { 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; if (dcmp (delta) == 0) { t1 = t2 = -f / (2 * e); P.push_back (L.point (t1)); return -1; } t1 = (-f - sqrt (delta)) / (2 * e); P.push_back (L.point (t1)); t2 = (-f + sqrt (delta)) / (2 * e); P.push_back (L.point (t2)); if (dcmp (t1) < 0 || dcmp (t2) < 0) return 0; return 2; } /* 两圆相交求交点,返回交点个数。交点保存在P中 */ int cir_cir_inter(Circle C1, Circle C2, vector<Point> &P) { double d = length (C1.c - C2.c); if (dcmp (d) == 0) { if (dcmp (C1.r - C2.r) == 0) return -1; //两圆重叠 else 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 = polar_angle (C2.c - C1.c); double da = acos ((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d)); //C1C2到C1P1的角? Point p1 = C1.point (a - da), p2 = C2.point (a + da); P.push_back (p1); if (p1 == p2) return 1; else P.push_back (p2); return 2; } /* 过点到圆的切线,返回切线条数,切线保存在V中 */ int point_cir_tan(Point p, Circle C, Vector *V) { Vector u = C.c - p; double dis = length (u); if (dis < C.r) return 0; else if (dcmp (dis - C.r) == 0) { V[0] = rotate (u, PI / 2); return 1; } else { double ang = asin (C.r / dis); V[0] = rotate (u, -ang); V[1] = rotate (u, +ang); return 0; } } /* 两圆的公切线,返回公切线条数,切线短点保存在a和b中 */ int cir_cir_tan(Circle A, Circle B, Point *a, Point *b) { int cnt = 0; if (A.r < B.r) { swap (A, B); swap (a, b); } double d = dot (A.c - B.c, A.c - B.c); double rsub = A.r - B.r, rsum = A.r + B.r; if (dcmp (d - rsub) < 0) return 0; //内含 double base = polar_angle (B.c - A.c); if (dcmp (d) == 0 && dcmp (A.r - B.r) == 0) return -1; //两圆重叠 if (dcmp (d - rsub) == 0) { //内切,一条切线 a[cnt] = A.point (base); b[cnt] = B.point (base); cnt++; return 1; } //有外公切线 double ang = acos (rsub / d); a[cnt] = A.point (base + ang); b[cnt] = B.point (base + ang); cnt++; a[cnt] = A.point (base - ang); b[cnt] = B.point (base - ang); cnt++; if (d == rsum) { a[cnt] = A.point (base); b[cnt] = B.point (base + PI); cnt++; } else if (dcmp (d - rsum) > 0) { //两条内公切线 double ang2 = acos (rsum / d); a[cnt] = A.point (base + ang2); b[cnt] = B.point (base + ang2 + PI); cnt++; a[cnt] = A.point (base - ang2); b[cnt] = B.point (base - ang2 + PI); cnt++; } return cnt; } Point p1[22], p2[22], p3; int main(void) { int n; while (scanf ("%d", &n) == 1) { if (!n) break; for (int i=1; i<=n; ++i) { p1[i] = read_point (); p2[i] = Point (p1[i].x, p1[i].y - 1); } bool flag = false; double ans = p1[1].x; for (int i=1; i<=n && !flag; ++i) { for (int j=1; j<=n && !flag; ++j) { if (i == j) continue; int k; for (k=1; k<=n; ++k) { if (!can_line_seg_inter (p1[i], p2[j], p1[k], p2[k])) { break; } } if (k == n + 1) { flag = true; break; } else if (k > max (i, j)) { p3 = line_line_inter (p1[i], p2[j] - p1[i], p1[k-1], p1[k] - p1[k-1]); ans = max (ans, p3.x); p3 = line_line_inter (p1[i], p2[j] - p1[i], p2[k-1], p2[k] - p2[k-1]); ans = max (ans, p3.x); } } } if (!flag) { printf ("%.2f\n", ans); } else puts ("Through all the pipe."); } //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"; return 0; }
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原文地址:http://www.cnblogs.com/Running-Time/p/4925387.html