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http://poj.org/problem?id=3384
题意:给一个凸包,求往里面放两个圆(可重叠)的最大面积时的两个圆心坐标。
思路:先把凸包边往内推R,做半平面交,然后做旋转卡壳,此时得到最大距离的点对,就是圆心坐标。
PS:最大长度的初始值要设置为负数,因为距离有可能退化到0,就像这组数据
4 1
0 0
2 0
2 2
0 2
#include<cstdio> #include<iostream> #include<cmath> #include<cstring> #include<algorithm> const double Pi=acos(-1); double R; int n,tot; struct Point{ double x,y; Point(){} Point(double x0,double y0):x(x0),y(y0){} }p[200005]; struct Line{ Point s,e; double slop; Line(){} Line(Point s0,Point e0):s(s0),e(e0){} }L[200005],l[200005],c[200005]; int read(){ int t=0,f=1;char ch=getchar(); while (ch<‘0‘||ch>‘9‘){if (ch==‘-‘)f=-1;ch=getchar();} while (‘0‘<=ch&&ch<=‘9‘){t=t*10+ch-‘0‘;ch=getchar();} return t*f; } Point operator /(Point p1,double x){ return Point(p1.x/x,p1.y/x); } Point operator *(Point p,double x){ return Point(p.x*x,p.y*x); } double operator *(Point p1,Point p2){ return p1.x*p2.y-p1.y*p2.x; } Point operator -(Point p1,Point p2){ return Point(p1.x-p2.x,p1.y-p2.y); } Point operator +(Point p1,Point p2){ return Point(p1.x+p2.x,p1.y+p2.y); } bool cmp(Line p1,Line p2){ if (p1.slop!=p2.slop) return p1.slop<p2.slop; else return (p1.e-p1.s)*(p2.e-p1.s)<=0; } Point inter(Line p1,Line p2){ double k1=(p2.e-p1.s)*(p1.e-p1.s); double k2=(p1.e-p1.s)*(p2.s-p1.s); double t=(k2)/(k1+k2); double x=p2.s.x+(p2.e.x-p2.s.x)*t; double y=p2.s.y+(p2.e.y-p2.s.y)*t; return Point(x,y); } bool jud(Line p1,Line p2,Line p3){ Point p=inter(p1,p2); return (p-p3.s)*(p3.e-p3.s)>0; } void phi(){ std::sort(l+1,l+1+tot,cmp); int cnt=1; for (int i=2;i<=tot;i++) if (l[i].slop!=l[i-1].slop) l[++cnt]=l[i]; int L=1,R=2;c[L]=l[1];c[R]=l[2]; for (int i=3;i<=cnt;i++){ while (L<R&&jud(c[R],c[R-1],l[i])) R--; while (L<R&&jud(c[L],c[L+1],l[i])) L++; c[++R]=l[i]; } while (L<R&&jud(c[R],c[R-1],c[L])) R--; while (L<R&&jud(c[L],c[L+1],c[R])) L++; tot=0; c[R+1]=c[L]; for (int i=L;i<=R;i++) p[++tot]=inter(c[i],c[i+1]); } double sqr(double x){ return x*x; } double dis(Point p){ return sqrt(sqr(p.x)+sqr(p.y)); } Point turn(Point p,double ang){ double Cos=cos(ang),Sin=sin(ang); double x=Cos*p.x-Sin*p.y; double y=Cos*p.y+Sin*p.x; return Point(x,y); } Point e(Point p){ double len=dis(p);p=p/len;return p; } double dis(Point p1,Point p2){ return dis(p1-p2); } void rc(){ p[tot+1]=p[1]; int k=2; double mx=-1; Point ans1,ans2; for (int i=1;i<=tot;i++){ while (fabs((p[i%tot+1]-p[i])*(p[k]-p[i]))<fabs((p[i%tot+1]-p[i])*(p[k%tot+1]-p[i]))) k=(k)%tot+1; if (mx<dis(p[i],p[k])){ mx=dis(p[i],p[k]); ans1=p[i]; ans2=p[k]; } } printf("%.4f %.4f %.4f %.4f",ans1.x,ans1.y,ans2.x,ans2.y); } int main(){ n=read(),R=read(); for (int i=1;i<=n;i++) p[i].x=read(),p[i].y=read(); for (int i=1;i<=n/2;i++) std::swap(p[i],p[n-i+1]); p[n+1]=p[1]; for (int i=1;i<=n;i++) l[++tot]=Line(p[i],p[i+1]),l[tot].slop=atan2(l[tot].e.y-l[tot].s.y,l[tot].e.x-l[tot].s.x); for (int i=1;i<=n;i++){ Point p=e(turn((l[i].e-l[i].s),Pi/2.0))*R; l[i].s=l[i].s+p; l[i].e=l[i].e+p; } phi(); rc(); }
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原文地址:http://www.cnblogs.com/qzqzgfy/p/5657500.html