hdu 4454 Stealing a Cake（三分法）

 

hdu 4454 Stealing a Cake（三分法）

分类： 乱搞 2013-09-01 17:28  219人阅读  评论(0)  收藏  举报

给定一个起始点，一个矩形，一个圆，三者互不相交。求从起始点->圆->矩形的最短距离。

自己画一画就知道距离和会是凹函数，不过不是一个凹函数。按与水平向量夹角为圆心角求圆上某点坐标，[0, PI] , [PI, 2*pi]两个区间的点会有两个凹函数。所以要做两次三分才行。

 #include<algorithm>
 #include<iostream>
 #include<fstream>
 #include<sstream>
 #include<cstring>
 #include<cstdlib>
 #include<string>
 #include<vector>
 #include<cstdio>
 #include<queue>
 #include<stack>
 #include<cmath>
 #include<map>
 #include<set>
 #define FF(i, a, b) for(int i=a; i<b; i++)
 #define FD(i, a, b) for(int i=a; i>=b; i--)
 #define REP(i, n) for(int i=0; i<n; i++)
 #define CLR(a, b) memset(a, b, sizeof(a))
 #define LL long long
 #define PB push_back
 #define eps 1e-10
 #define debug puts("**debug**");
 using namespace std;
 const double PI = acos(-1);

 struct Point
 {
     double x, y;
     Point(double x=0, double y=0):x(x), y(y){}
 };
 typedef Point Vector;

 Vector operator + (Vector a, Vector b) { return Vector(a.x+b.x, a.y+b.y); }
 Vector operator - (Vector a, Vector b) { return Vector(a.x-b.x, a.y-b.y); }
 Vector operator * (Vector a, double p) { return Vector(a.x*p, a.y*p); }
 Vector operator / (Vector a, double p) { return Vector(a.x/p, a.y/p); }
 bool operator < (const Point& a, const Point& b)
 {
     return a.x < b.x || (a.x == b.x && a.y < b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x) < eps) return 0; return x < 0 ? -1 : 1;
 }
 bool operator == (const Point& a, const Point& b)
 {
     return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
 }

 double Dot(Vector a, Vector b) { return a.x*b.x + a.y*b.y; }
 double Length(Vector a) { return sqrt(Dot(a, a)); }
 double Cross(Vector a, Vector b) { return a.x*b.y - a.y*b.x; }
 double DistanceToSegment(Point p, Point a, Point b)
 {
     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);
 }
 struct Circle
 {
     Point c;
     double r;
     Circle(){}
     Circle(Point c, double r):c(c), r(r){}
     Point point(double a) //根据圆心角求点坐标
     {
         return Point(c.x+cos(a)*r, c.y+sin(a)*r);
     }
 }o;

 Point p, p1, p2, p3, p4, s;
 double a, b, c, d;


 double Calc(double x)
 {
     p = o.point(x);
     double d1 = DistanceToSegment(p, p1, p2),
     d2 = DistanceToSegment(p, p2, p3),
     d3 = DistanceToSegment(p, p3, p4),
     d4 = DistanceToSegment(p, p4, p1);
     //点p到矩形最近距离加上s到p距离
     return min(min(d1, d2), min(d3, d4)) + Length(s-p);
 }

 double solve()
 {
     double L, R, m, mm, mv, mmv;
     L = 0; R = PI;
     while (L + eps < R)
     {
         m = (L + R) / 2;
         mm = (m + R) / 2;
         mv = Calc(m);
         mmv = Calc(mm);
         if (mv <= mmv) R = mm; //三分法求最大值时改为mv>=mmv
         else L = m;
     }
     double ret = Calc(L);
     L = PI; R = 2*PI;
     while (L + eps < R)
     {
         m = (L + R) / 2;
         mm = (m + R) / 2;
         mv = Calc(m);
         mmv = Calc(mm);
         if (mv <= mmv) R = mm;
         else L = m;
     }
     return min(ret, Calc(L));
 }

 int main()
 {
     while(scanf("%lf%lf", &s.x, &s.y))
     {
         if(s.x == 0 && s.y == 0) break;
         scanf("%lf%lf%lf", &o.c.x, &o.c.y, &o.r);
         scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
         //确定矩形四个点坐标，左上点开始 逆时针
         double maxx, maxy, minx, miny;
         maxx = max(a, c); maxy = max(b, d);
         minx = min(a, c); miny = min(b, d);
         p1 = Point(minx, maxy);
         p2 = Point(minx, miny);
         p3 = Point(maxx, miny);
         p4 = Point(maxx, maxy);
         double ans = solve();
         printf("%.2f\n", ans);
     }
     return 0;
 }