2018 ACM-ICPC, Syrian Collegiate Programming Contest I. Rise of the Robots 最小圆覆盖模板

题目链接：http://codeforces.com/gym/102006/problem/I

题意：

一个机器人要移动n次，每次移动的路径已知，告诉你桌子的大小和机器人的大小，问你不让机器人移动出桌子的初始位置在哪里。

做法：

因为没有完全确定就不说了留个板子就好、

#include<bits/stdc++.h>
using namespace std;
const double eps=1e-6;
const double Pi=acos(-1.0);
const int maxn=255;
struct node{double x,y;node(){}node(double x,double y):x(x),y(y){}
} P[maxn],c;
int n;
typedef node Vector;
int dcmp(double x)
{if (fabs(x)<eps) return 0;else if (x<0) return -1;else return 1;
}
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 node &a, const node &b){return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}
double Dot(Vector a, Vector b){//内积return a.x*b.x + a.y*b.y;
}
double Length(Vector a){//模return sqrt(Dot(a, a));
}
double Angle(Vector a, Vector b){//夹角,弧度制return acos(Dot(a, b) / Length(a) / Length(b));
}
double Cross(Vector a, Vector b){//外积return a.x*b.y - a.y*b.x;
}
Vector Rotate(Vector a, double rad){//逆时针旋转return Vector(a.x*cos(rad) - a.y*sin(rad), a.x*sin(rad) + a.y*cos(rad));
}
double Distance(node a, node b){//两点间距离return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
}
double Area(node a, node b, node c){//三角形面积return fabs(Cross(b - a, c - a) / 2);
}double lenth(node a) {return sqrt(Dot(a,a));}node rotate(node a,double t)    //向量旋转
{return node(a.x*cos(t)-a.y*sin(t),a.x*sin(t)+a.y*cos(t));
}node jiao(node p,node v,node q,node w)
//p+tv
//q+tw
{node u=p-q;double t=Cross(w,u)/Cross(v,w);return p+v*t;
}node get_c(node a,node b,node c)
{node p=(a+b)/2;    //ad中点node q=(a+c)/2;    //ac中点node v=rotate(b-a,Pi/2.0),w=rotate(c-a,Pi/2.0);   //中垂线的方向向量if (dcmp(Cross(v,w))==0)    //平行{if (dcmp(Length(a-b)+Length(b-c)-Length(a-c))==0)return (a+c)/2;if (dcmp(Length(b-a)+Length(a-c)-Length(b-c))==0)return (b+c)/2;if (dcmp(Length(a-c)+Length(c-b)-Length(a-b))==0)return (a+b)/2;}return jiao(p,v,q,w);
}void min_circular(){random_shuffle(P+1,P+n+1);    //随机化double r;c=P[1],r=0;//c 圆心//r 半径for (int i=2;i<=n;i++)if (dcmp(lenth(c-P[i])-r)>0)    //不在圆内{c=P[i],r=0;for (int j=1;j<i;j++)if (dcmp(lenth(c-P[j])-r)>0){c=(P[i]+P[j])/2.0;r=lenth(c-P[i]);for (int k=1;k<j;k++)if (dcmp(lenth(c-P[k])-r)>0){c=get_c(P[i],P[j],P[k]);r=lenth(c-P[i]);}}}
}
int main(){freopen("robots.in","r",stdin);int t;cin>>t;while(t--){double R,r,xx,yy;scanf("%d%lf%lf",&n,&R,&r);P[1]=node(0,0);n++;for(int i=2;i<=n;i++){scanf("%lf%lf",&xx,&yy);P[i]=P[i-1]+node(xx,yy);}min_circular();printf("%.8f %.8f\n",-c.x,-c.y);}
}

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