题意:凸多边形的小岛在海里,问岛上的点到海最远的距离。
分析:训练指南P279,二分答案,然后整个多边形往内部收缩,如果半平面交非空,那么这些点构成半平面,存在满足的点。
/************************************************ * Author :Running_Time * Created Time :2015/11/10 星期二 14:16:17 * File Name :LA_3890.cpp ************************************************/ #include <bits/stdc++.h> 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 length(Vector A) { //向量长度,点积 return sqrt (dot (A, A)); } double polar_angle(Vector A) { //向量极角 return atan2 (A.y, A.x); } Vector nomal(Vector A) { //向量的单位法向量 double len = length (A); return Vector (-A.y / len, A.x / len); } struct Line { Point p; Vector v; double ang; Line () {} Line (const Point &p, const Vector &v) : p (p), v (v) { ang = polar_angle (v); } bool operator < (const Line &r) const { return ang < r.ang; } Point point(double a) { return p + v * a; } }; bool point_on_left(Point p, Line L) { return cross (L.v, p - L.p) > 0; } 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; } vector<Point> half_plane_inter(vector<Line> L) { sort (L.begin (), L.end ()); int first, last, n = L.size (); Point *p = new Point[n]; Line *q = new Line[n]; q[first=last=0] = L[0]; for (int i=1; i<n; ++i) { while (first < last && !point_on_left (p[last-1], L[i])) last--; while (first < last && !point_on_left (p[first], L[i])) first++; q[++last] = L[i]; if (dcmp (cross (q[last].v, q[last-1].v)) == 0) { last--; if (point_on_left (L[i].p, q[last])) q[last] = L[i]; } if (first < last) p[last-1] = line_line_inter (q[last-1].p, q[last-1].v, q[last].p, q[last].v); } while (first < last && !point_on_left (p[last-1], q[first])) last--; vector<Point> ps; if (last - first <= 1) return ps; p[last] = line_line_inter (q[last].p, q[last].v, q[first].p, q[first].v); for (int i=first; i<=last; ++i) ps.push_back (p[i]); return ps; } Point ps[110]; Vector V[110], V2[110]; int main(void) { int n; while (scanf ("%d", &n) == 1) { if (!n) break; for (int i=0; i<n; ++i) ps[i] = read_point (); for (int i=0; i<n; ++i) { V[i] = ps[(i+1)%n] - ps[i]; V2[i] = nomal (V[i]); } double l = 0, r = 20000; while (r - l > EPS) { double mid = l + (r - l) / 2; vector<Line> L; for (int i=0; i<n; ++i) { L.push_back (Line (ps[i] + V2[i] * mid, V[i])); } vector<Point> qs = half_plane_inter (L); int sz = qs.size (); if (sz == 0) r = mid; else l = mid; } printf ("%.6f ", l); } //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s. "; return 0; }