最小覆盖双圆问题 | Kyons'Blog 




 
 
 
 



 
 



最小覆盖双圆问题

前置知识:最小圆覆盖

洛谷P1742 最小圆覆盖
  给定平面一定数量的点,找一个最小的圆,将所有点包含进去.
  然后,就不记得是哪位大佬提出了一种$O(n)$的解决方式,随机增量法.

  1. 保证数据的随机性,我们使用$random\_shuffle$这个函数
  2. 我们要接受一个定理:如果点$p$不在集合$S$的最小覆盖圆内,那么肯定在$p\cup S$的最小覆盖圆上.
  3. 首先,假设我们已经求出了前$i$个点形成的最小覆盖圆
  4. 如果$i+1$这个点不在$C_i$圆内,那么该$P_{i+1}$必定在$C_{i+1}$上
  5. 那么我们以$P_{i+1}$为圆心,半径为$0$作为起始圆,找到第$j$个点,保证$dis\{i+1,j\}>$当前圆的半径.以这两点为圆$C_{i+1,j}$半径为$dis\{i+1,j\}$,圆心为$\frac{P_{i+1}+P_j}{2}$
  6. 再寻找第三个点不在$C_{i+1,j}$,设为第$k$个点,那么以$i,j,k$三点的圆即可被认为是期望的最小覆盖圆$C_{i+1,j,k}$
  7. 不断重复$3-6$步骤,即可得到答案.

  体感就是$O(n^3)$的嘛,在数据不随机的情况下,会退化为$O(n^3)$,但在数据完全随机的情况下,预期期望复杂度为$O(n)$.

代码戳我戳我
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cirles zuixiaoyuan_fugai(int l, int r) //最小圆覆盖
{
if (l > r)
return 0;
for (int i = l; i <= r; i++)
s[i] = p[i];

random_shuffle(s + l, s + r + 1);
cirles c(s[l], 0.0);
for (int i = l; i <= r; i++)
if (zhengfu(sqr(c.o - s[i]) - c.r) > 0) {
c = cirles(s[i], 0.0);
for (int j = l; j < i; j++)
if (zhengfu(sqr(c.o - s[j]) - c.r) > 0) {
c.o = (s[i] + s[j]) * 0.5, c.r = sqr(c.o - s[j]);
for (int k = l; k < j; k++)
if (zhengfu(sqr(c.o - s[k]) - c.r) > 0) {
c.o = sanjiaoxing_waixin(s[i], s[j], s[k]), c.r = sqr(c.o - s[k]);
}
}
}
return c;
}

最小覆盖双圆问题

洛谷P4586 最小覆盖双圆问题
又是一道考验人类智慧的问题.
首先,我们考虑最好情况.及两圆半径相同,均分了两侧的点.
如下图:(大概的画画就行啦)
1.png
反正就是如何让两边的圆的半径接近相同.
很显然的二分,先对两边各做一次最小圆,得到两个圆的半径,我们只需要在大的半径里二分一哈.
然后就可以得到答案啦.
但是有个问题,划分两个集合的这条线可能是斜的,所以我们还要暴力转角.
每次转那么$1°$,转那么$180$次就差不多.
转的越多,精度越高,时间越长.(只要相信你自己是欧皇,你就可以不超时又$AC$)

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#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
using namespace std;
const int MAXN = 2 * 1e3 + 5;
const double INF = 1e18;
const double alpha = 1.0 / 180 * acos(-1);
const double eps = 1e-9;
const double pi = acos(-1.0);

typedef struct point vec; //向量vec
struct point { //点的基本数据结构
double x, y;
point(double _x = 0, double _y = 0)
: x(_x)
, y(_y)
{
}
point operator*(const point i_T) const
{
return point(x * i_T.x, y * i_T.y);
}
point operator*(double u) const
{
return point(x * u, y * u);
}
bool operator==(const point i_T) const
{
return x == i_T.x && y == i_T.y;
}
point operator/(double u) const
{
return point(x / u, y / u);
}
point operator+(const point i_T)
{
return point(x + i_T.x, y + i_T.y);
}
point operator-(const point i_T)
{
return point(x - i_T.x, y - i_T.y);
}
friend bool operator<(point a, point b)
{
return a.x == b.x ? a.y < b.y : a.x < b.x;
}
friend ostream& operator<<(ostream& out, point& a)
{
//cout << a.x << ' ' << a.y;
printf("%.8f %.8f", a.x, a.y);
return out;
}
friend istream& operator>>(istream& in, point& a)
{
scanf("%lf%lf", &a.x, &a.y);
return in;
}
};
struct cirles {
point o;
double r;
cirles(point _o = point(), double _r = 0.0)
: r(_r)
, o(_o)
{
}
point Point(double t) //圆上任意一点
{
return point(o.x + r * cos(t), o.y + r * sin(t));
}
friend istream& operator>>(istream& in, cirles& a)
{
in >> a.o >> a.r;
return in;
}
friend ostream& operator<<(ostream& out, cirles& a)
{
out << a.o << ' ' << a.r;
return out;
}
};
//求圆心
point s[MAXN], p[MAXN];
int zhengfu(double d) //判断正负
{
if (fabs(d) < eps)
return 0;
if (d > 0)
return 1;
return -1;
}
double changdu(vec a)
{
return sqrt(a.x * a.x + a.y * a.y);
}
double sqr(vec a)
{
return a.x * a.x + a.y * a.y;
}
point sanjiaoxing_waixin(point a, point b, point c) //三角形外心
{
double A = a.x * a.x - b.x * b.x + a.y * a.y - b.y * b.y, B = (b.x - a.x) * 2, C = (b.y - a.y) * 2;
double D = c.x * c.x - b.x * b.x + c.y * c.y - b.y * b.y, E = (b.x - c.x) * 2, F = (b.y - c.y) * 2;
return point((D * C - A * F) / (B * F - E * C), (D * B - A * E) / (C * E - F * B));
}
vec xiangliang_xuanzhuan(vec a, double k) //将向量按照起点旋转逆时针k,
{
double x, y;
x = cos(k) * a.x - sin(k) * a.y;
y = sin(k) * a.x + cos(k) * a.y;
return vec(x, y);
}
double zuixiaoyuan_fugai(int l, int r) //最小圆覆盖
{
if (l > r)
return 0;
for (int i = l; i <= r; i++)
s[i] = p[i];

random_shuffle(s + l, s + r + 1);
cirles c(s[l], 0.0);
for (int i = l; i <= r; i++)
if (zhengfu(sqr(c.o - s[i]) - c.r) > 0) {
c = cirles(s[i], 0.0);
for (int j = l; j < i; j++)
if (zhengfu(sqr(c.o - s[j]) - c.r) > 0) {
c.o = (s[i] + s[j]) * 0.5, c.r = sqr(c.o - s[j]);
for (int k = l; k < j; k++)
if (zhengfu(sqr(c.o - s[k]) - c.r) > 0) {
c.o = sanjiaoxing_waixin(s[i], s[j], s[k]), c.r = sqr(c.o - s[k]);
}
}
}
return c.r;
}
int main()
{
//freopen("txt.txt", "w", stdout);
srand(20030719);
int n;
while (scanf("%d", &n) && n) {
for (int i = 1; i <= n; i++)
cin >> p[i];
double ans = INF;
for (int i = 1; i <= 181; i++) {
sort(p + 1, p + 1 + n);
int l = 1, r = n;
while (l <= r) {
int mid = (l + r) >> 1;
double retl = zuixiaoyuan_fugai(1, mid);
double retr = zuixiaoyuan_fugai(mid + 1, n);
double tmp = max(retl, retr);
if (retr + retl - tmp > ans)
break;
ans = min(tmp, ans);
if (retl > retr)
r = mid - 1;
else
l = mid + 1;
}
for (int i = 1; i <= n; i++)
p[i] = xiangliang_xuanzhuan(p[i], alpha);
}
printf("%.2lf\n", sqrt(ans));
}
return 0;
}

 


 
 
 





 

 


 


 

 

 

 

 
 

 

 

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