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

时间:2014-10-21 11:52:38      阅读:152      评论:0      收藏:0      [点我收藏+]

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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。

 

bubuko.com,布布扣
  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

标签:style   blog   http   color   io   os   ar   for   sp   

原文地址:http://www.cnblogs.com/AOQNRMGYXLMV/p/4039878.html

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