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  • poj 3384 Feng Shui (Half Plane Intersection)

    3384 -- Feng Shui

      构造半平面交,然后求凸包上最远点对。

      这题的题意是给出一个凸多边形区域,要求在其中放置两个半径为r的圆(不能超出凸多边形区域),要求求出两个圆心,使得多边形中没有被覆盖的面积最小。反之就是求圆覆盖的区域最大。首先我们可以求出圆心放置的位置的区域,这个要利用半平面交,将原多边形区域向内收缩r的距离。要求两个圆覆盖的区域最大,也就是它们相交的面积最小,也就是两个圆心的距离要尽可能的大。这样就说明了,这题的做法是要求出凸包上面的最远点对。

      做这题的时候犯了两个错误,一个是没有设置对精度,直接用了cout的默认输出,另一个则是没有想到收缩以后,剩余的多边形的顶点数会少于n。

    代码如下:

      1 #include <cstdio>
      2 #include <cstring>
      3 #include <iostream>
      4 #include <algorithm>
      5 #include <vector>
      6 #include <cmath>
      7 
      8 using namespace std;
      9 
     10 struct Point {
     11     double x, y;
     12     Point() {}
     13     Point(double x, double y) : x(x), y(y) {}
     14 } ;
     15 template<class T> T sqr(T x) { return x * x;}
     16 
     17 typedef Point Vec;
     18 Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);}
     19 Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);}
     20 Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);}
     21 Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);}
     22 
     23 const double EPS = 1e-8;
     24 const double PI = acos(-1.0);
     25 inline int sgn(double x) { return (x > EPS) - (x < -EPS);}
     26 bool operator < (Point a, Point b) { return sgn(a.x - b.x) < 0 || sgn(a.x - b.x) == 0 && a.y < b.y;}
     27 bool operator == (Point a, Point b) { return sgn(a.x - b.x) == 0 || sgn(a.y - b.y) == 0;}
     28 
     29 inline double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;}
     30 inline double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;}
     31 inline double dotDet(Point o, Point a, Point b) { return dotDet(a - o, b - o);}
     32 inline double crossDet(Point o, Point a, Point b) { return crossDet(a - o, b - o);}
     33 inline double vecLen(Vec x) { return sqrt(dotDet(x, x));}
     34 inline double toRad(double deg) { return deg/ 180.0 * PI;}
     35 inline double angle(Vec v) { return atan2(v.y, v.x);}
     36 Vec normal(Vec x) {
     37     double len = vecLen(x);
     38     return Vec(-x.y, x.x) / len;
     39 }
     40 
     41 struct Poly {
     42     vector<Point> pt;
     43     Poly() { pt.clear();}
     44     ~Poly() {}
     45     Poly(vector<Point> &pt) : pt(pt) {}
     46     Point operator [] (int x) const { return pt[x];}
     47     int size() { return pt.size();}
     48     double area() {
     49         double ret = 0.0;
     50         int sz = pt.size();
     51         for (int i = 1; i < sz; i++) {
     52             ret += crossDet(pt[i], pt[i - 1]);
     53         }
     54         return fabs(ret / 2.0);
     55     }
     56 } ;
     57 
     58 struct DLine {
     59     Point p;
     60     Vec v;
     61     double ang;
     62     DLine() {}
     63     DLine(Point p, Vec v) : p(p), v(v) { ang = atan2(v.y, v.x);}
     64     bool operator < (const DLine &L) const { return ang < L.ang;}
     65     DLine move(double x) {
     66         Vec nor = normal(v);
     67         nor = nor * x;
     68         return DLine(p + nor, v);
     69     }
     70 } ;
     71 
     72 inline bool onLeft(DLine L, Point p) { return crossDet(L.v, p - L.p) > 0;}
     73 Point dLineIntersect(DLine a, DLine b) {
     74     Vec u = a.p - b.p;
     75     double t = crossDet(b.v, u) / crossDet(a.v, b.v);
     76     return a.p + a.v * t;
     77 }
     78 
     79 Poly halfPlane(DLine *L, int n) {
     80     Poly ret = Poly();
     81     sort(L, L + n);
     82     int fi, la;
     83     Point *p = new Point[n];
     84     DLine *q = new DLine[n];
     85     q[fi = la = 0] = L[0];
     86     for (int i = 1; i < n; i++) {
     87         while (fi < la && !onLeft(L[i], p[la - 1])) la--;
     88         while (fi < la && !onLeft(L[i], p[fi])) fi++;
     89         q[++la] = L[i];
     90         if (fabs(crossDet(q[la].v, q[la - 1].v)) < EPS) {
     91             la--;
     92             if (onLeft(q[la], L[i].p)) q[la] = L[i];
     93         }
     94         if (fi < la) p[la - 1] = dLineIntersect(q[la - 1], q[la]);
     95     }
     96     while (fi < la && !onLeft(q[fi], p[la - 1])) la--;
     97     if (la < fi) return ret;
     98     p[la] = dLineIntersect(q[la], q[fi]);
     99     for (int i = fi; i <= la; i++) ret.pt.push_back(p[i]);
    100     return ret;
    101 }
    102 
    103 const int N = 111;
    104 Point pt[N];
    105 DLine dl[N];
    106 
    107 int main() {
    108 //    freopen("in", "r", stdin);
    109     int n;
    110     double r;
    111     while (cin >> n >> r) {
    112         for (int i = 0; i < n; i++) {
    113             cin >> pt[i].x >> pt[i].y;
    114             if (i) dl[i - 1] = DLine(pt[i], pt[i - 1] - pt[i]).move(r + EPS);
    115         }
    116         dl[n - 1] = DLine(pt[0], pt[n - 1] - pt[0]).move(r + EPS);
    117         Poly tmp = halfPlane(dl, n);
    118         if (tmp.size() <= 1) {
    119             for (int i = 0; i < n; i++) {
    120                 if (i) dl[i - 1] = DLine(pt[i], pt[i - 1] - pt[i]).move(r - EPS);
    121             }
    122             dl[n - 1] = DLine(pt[0], pt[n - 1] - pt[0]).move(r - EPS);
    123             tmp = halfPlane(dl, n);
    124         }
    125         double dis = 0.0;
    126         int id[2] = { 0, 0};
    127         n = tmp.size();
    128         for (int i = 0; i < n; i++) {
    129             for (int j = 0; j < n; j++) {
    130                 if (dis < vecLen(tmp[i] - tmp[j])) {
    131                     dis = vecLen(tmp[i] - tmp[j]);
    132                     id[0] = i;
    133                     id[1] = j;
    134                 }
    135             }
    136         }
    137 //        cout << vecLen(tmp[id[0]] - tmp[id[1]]) << endl;
    138         cout.precision(9);
    139         cout << tmp[id[0]].x << ' ' << tmp[id[0]].y << ' ' << tmp[id[1]].x << ' ' << tmp[id[1]].y << endl;
    140     }
    141     return 0;
    142 }
    View Code

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    ——written by Lyon

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  • 原文地址:https://www.cnblogs.com/LyonLys/p/poj_3384_Lyon.html
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