• bzoj 2823: [AHOI2012]信号塔 最小圆覆盖


    题目大意:

    给定n个点,求面积最小的园覆盖所有点.其中(n leq 10^6)

    题解:

    恩。。。
    刚拿到这道题的时候...

    什么???最小圆覆盖不是(O(n^3))的随机增量算法吗?????
    (10^6)又是个什么鬼?????????

    然后去%了popoqqq大爷的题解...原来这道题数据是随机的啊。。。
    随机数据有一个性质,在凸包上的点不超过(logn)
    所以我们求凸包然后在上面跑随机增量算法即可

    #include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    typedef long long ll;
    inline void read(int &x){
    	x=0;char ch;bool flag = false;
    	while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
    	while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
    }
    inline int cat_max(const int &a,const int &b){return a>b ? a:b;}
    inline int cat_min(const int &a,const int &b){return a<b ? a:b;}
    const int maxn = 1000010;
    const double eps = 1e-9;
    inline int dcmp(const double &x){
    	if(x < eps && x > -eps) return 0;
    	return x > 0 ? 1 : -1;
    }
    struct Point{
    	double x,y;
    	Point(const double &a = 0,const double &b = 0){x=a;y=b;}
    	void print(){
    		printf("Point : (%lf,%lf)
    ",x,y);
    	}
    };
    typedef Point Vector;
    inline Vector operator + (const Vector &a,const Vector &b){
    	return Vector(a.x+b.x,a.y+b.y);
    }
    inline Vector operator - (const Vector &a,const Vector &b){
    	return Vector(a.x-b.x,a.y-b.y);
    }
    inline Vector operator / (const Vector &a,const double &b){
    	return Vector(a.x/b,a.y/b);
    }
    inline bool operator < (const Point &a,const Point &b){
    	return a.x == b.x ? a.y < b.y : a.x < b.x;
    }
    inline double operator * (const Vector &a,const Vector &b){
    	return a.x*b.x + a.y*b.y;
    }
    inline double cross(const Vector &a,const Vector &b){
    	return a.x*b.y - a.y*b.x;
    }
    inline double sqr(const double &x){
    	return x*x;
    }
    inline double dis(const Point &a,const Point &b){
    	return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
    }
    inline Point getPoint(const Point &p0,const Point &p1,const Point &p2){
    	double a1 = p1.x - p0.x,b1 = p1.y - p0.y,c1 = (a1*a1 + b1*b1)/2.0;
    	double a2 = p2.x - p0.x,b2 = p2.y - p0.y,c2 = (a2*a2 + b2*b2)/2.0;
    	double d = a1*b2 - a2*b1;
    	return Point(p0.x+(c1*b2-c2*b1)/d,p0.y+(a1*c2-a2*c1)/d);
    } 
    Point o;double r;
    inline bool in_cir(const Point &p){
    	return dcmp(dis(p,o) - r) <= 0;
    }
    Point p[maxn];int n;
    inline void getMinCir(){
    	o = Point(0,0);r = 0;
    	for(int i=2;i<=n;++i){ 
    		if(!in_cir(p[i])){ o = p[i];r = 0;
    			for(int j=1;j<i;++j){ 
    				if(!in_cir(p[j])){o = (p[i]+p[j])/2.0;r = dis(p[i],o);
    					for(int k=1;k<j;++k){
    						if(!in_cir(p[k])){
    							o = getPoint(p[i],p[j],p[k]);
    							r = dis(p[i],o);
    						}
    					}
    				}
    			}
    		}
    	}return;
    }
    Point ch[maxn];int m;
    inline void convex(){
    	sort(p+1,p+n+1);m = 0;
    	for(int i=1;i<=n;++i){
    		while(m > 1 && cross(ch[m] - ch[m-1],p[i] - ch[m]) <= 0) -- m;
    		ch[++m] = p[i];
    	}int k = m;
    	for(int i=n-1;i>=1;--i){
    		while(m > k && cross(ch[m] - ch[m-1],p[i] - ch[m]) <= 0) -- m;
    		ch[++m] = p[i];
    	}if(n > 1) -- m;
    	swap(n,m);swap(p,ch);
    }
    int main(){
    	read(n);
    	for(int i=1;i<=n;++i){
    		scanf("%lf%lf",&p[i].x,&p[i].y);
    	}
    	convex();getMinCir();
    	printf("%.2lf %.2lf %.2lf
    ",o.x,o.y,r);
    	getchar();getchar();
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Skyminer/p/6435538.html
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