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LA 3890 (半平面交) Most Distant Point from the Sea

题意:

给出一个凸n边形,求多边形内部一点使得该点到边的最小距离最大。

分析:

最小值最大可以用二分。

多边形每条边的左边是一个半平面,将这n个半平面向左移动距离x,则将这个凸多边形缩小了。如果这n个半平面交非空,则存在这样距离为x的点,反之则不存在。

 

半平面交的代码还没有完全理解。

和凸包类似,先对这些半平面进行极角排序。每次新加入的平面可能让队尾的平面变得“无用”,从而需要删除。由于极角序是环形的,所以也可能让队首元素变得“无用”。所以用双端队列来维护。

 

  1 //#define LOCAL  2 #include <cstdio>  3 #include <cstring>  4 #include <algorithm>  5 #include <cmath>  6 #include <vector>  7 using namespace std;  8   9 const double eps = 1e-6; 10  11 struct Point 12 { 13     double x, y; 14     Point(double x=0, double y=0):x(x), y(y){} 15 }; 16 typedef Point Vector; 17 Point operator + (Point A, Point B) { return Point(A.x+B.x, A.y+B.y); } 18 Point operator - (Point A, Point B) { return Point(A.x-B.x, A.y-B.y); } 19 Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); } 20 Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } 21 double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } 22 double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } 23 double Length(Vector A) { return sqrt(Dot(A, A)); } 24 Vector Normal(Vector A) { double l = Length(A); return Vector(-A.y/l, A.x/l); } 25  26 double PolygonArea(const vector<Point>& p) 27 { 28     double ans = 0.0; 29     int n = p.size(); 30     for(int i = 1; i < n-1; ++i) 31         ans += Cross(p[i]-p[0], p[i+1]-p[0]); 32     return ans/2; 33 } 34  35 struct Line 36 { 37     Point p; 38     Vector v; 39     double ang; 40     Line() {} 41     Line(Point p, Vector v):p(p), v(v) { ang = atan2(v.y, v.x); } 42     bool operator < (const Line& L) const 43     { 44         return ang < L.ang; 45     } 46 }; 47  48 bool OnLeft(const Line& L, const Point& p) 49 { return Cross(L.v, p-L.p) > 0; } 50  51 Point GetLineIntersection(const Line& a, const Line& b) 52 { 53     Vector u = a.p-b.p; 54     double t = Cross(b.v, u) / Cross(a.v, b.v); 55     return a.p + a.v*t; 56 } 57  58 vector<Point> HalfplaneIntersection(vector<Line> L) 59 { 60     int n = L.size(); 61     sort(L.begin(), L.end()); 62  63     int first, last; 64     vector<Point> p(n); 65     vector<Line> q(n); 66     vector<Point> ans; 67      68     q[first=last=0] = L[0]; 69     for(int i = 1; i < n; ++i) 70     { 71         while(first < last && !OnLeft(L[i], p[last-1])) last--; 72         while(first < last && !OnLeft(L[i], p[first])) first++; 73         q[++last] = L[i]; 74         if(fabs(Cross(q[last].v, q[last-1].v)) < eps)    //Èç¹ûÁ½Ö±ÏßƽÐУ¬È¡ÄÚ²àµÄÄǸö  75         { 76             last--; 77             if(OnLeft(q[last], L[i].p)) q[last] = L[i]; 78         } 79         if(first < last) p[last-1] = GetLineIntersection(q[last-1], q[last]); 80     } 81     while(first < last && !OnLeft(q[first], p[last-1])) last--; 82     if(last - first <= 1)    return ans; 83     p[last] = GetLineIntersection(q[last], q[first]); 84      85     for(int i = first; i <= last; ++i)    ans.push_back(p[i]); 86     return ans; 87 } 88  89 int main(void) 90 { 91     #ifdef LOCAL 92         freopen("3890in.txt", "r", stdin); 93     #endif 94      95     int n; 96     while(scanf("%d", &n) == 1 && n) 97     { 98         vector<Point> p, v, normal; 99         int m, x, y;100         for(int i = 0; i < n; ++i) { scanf("%d%d", &x, &y); p.push_back(Point(x, y)); }101         if(PolygonArea(p) < 0) reverse(p.begin(), p.end());102         103         for(int i = 0; i < n; ++i)104         {105             v.push_back(p[(i+1)%n] - p[i]);106             normal.push_back(Normal(v[i]));107         }108         109         double left = 0, right = 20000;110         while(right - left > 5e-7)111         {112             vector<Line> L;113             double mid = (right + left) / 2;114             for(int i = 0; i < n; ++i) L.push_back(Line(p[i] + normal[i]*mid, v[i]));115             vector<Point> Poly = HalfplaneIntersection(L);116             if(Poly.empty())    right = mid;117             else left = mid;118         }119         printf("%.6lf\n", left);120     }121     122     return 0;123 }
代码君

 

LA 3890 (半平面交) Most Distant Point from the Sea