D.Country Meow 最小球覆盖 三分套三分套三分 && 模拟退火

Posted 2023-01-02 izcat

// 2019.10.3
// 练习题：2018 ICPC 南京现场赛

D Country Meow

题目大意

给定空间内 N 个点，求某个点到 N 个点的距离最大值的最小值。
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思路

非常裸的最小球覆盖问题啊，即找到半径最小的球包含全部的点。
在最小圆覆盖问题上，可以使用随机增量法，这里没有四点确定球心的公式，所以板子失效了。
最小圆覆盖可以用三分套三分，这里空间有三维，假装证明得到在任意一维上都满足凸函数特性，那么再套一层维度三分就OK了。
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AC代码

三分套三分套三分写法，复杂度O(n*log^3)

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps = 1e-3;
struct Point 
    double x, y, z;
    Point() 
        x = y = z = 0.0;
    
    Point(double xx, double yy, double zz) 
        x = xx, y = yy, z = zz;
    
    Point operator-(const Point& p) 
        return Point(x-p.x, y-p.y, z-p.z);
    
    double dis() 
        return sqrt(x*x+y*y+z*z);
    
pt[110];
int n;

double cal(double x, double y, double z) 
    double res = 0;
    for(int i=1;i<=n;i++) 
        res = max(res, (pt[i]-Point(x, y, z)).dis());
    
    return res;


double cal2(double x, double y) 
    double res = 1e18;
    double l = -100000, r = 100000;
    while(r-l>eps) 
        double m1 = (r-l)/3 + l;
        double m2 = (r-l)/3*2 + l;
        double res1 = cal(x, y, m1), res2 = cal(x, y, m2);
        res = min(res, min(res1, res2));
        if(res1<res2) r = m2;
        else l = m1;
    
    return res;


double cal3(double x) 
    double res = 1e18;
    double l = -100000, r = 100000;
    while(r-l>eps) 
        double m1 = (r-l)/3 + l;
        double m2 = (r-l)/3*2 + l;
        double res1 = cal2(x, m1), res2 = cal2(x, m2);
        res = min(res, min(res1, res2));
        if(res1<res2) r = m2;
        else l = m1;
    
    return res;


int main() 
    cin>>n;
    for(int i=1;i<=n;i++) 
        scanf("%lf %lf %lf", &pt[i].x, &pt[i].y, &pt[i].z);
    

    double res = 1e18;
    double l = -100000, r = 100000;
    while(r-l>eps) 
        double m1 = (r-l)/3 + l;
        double m2 = (r-l)/3*2 + l;
        double res1 = cal3(m1), res2 = cal3(m2);
        res = min(res, min(res1, res2));
        if(res1<res2) r = m2;
        else l = m1;
    
    printf("%.10lf\n", res);
    return 0;

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模拟退火写法，对于三维复杂度更低：

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps = 1e-5;
struct Point
    double x, y, z;
p[110], op;
int n;
 
inline double dist(Point &a, Point &b) 
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));

void solve() 
    double ans, delta = 10000.0;
    double maxDis, tempDis;
    while(delta>eps)
        int id = 0;
        maxDis = dist(op, p[id]);
        for(int i=1;i<n;i++)
            tempDis=dist(op,p[i]);
            if(tempDis>maxDis)
                maxDis = tempDis;
                id = i;
            
        
        ans = maxDis;
        op.x += (p[id].x-op.x)/maxDis*delta;
        op.y += (p[id].y-op.y)/maxDis*delta;
        op.z += (p[id].z-op.z)/maxDis*delta;
        delta *= 0.98;
    
    printf("%.10lf\n", ans);

 
int main() 
    while(scanf("%d", &n)!=EOF && n) 
        op.x = op.y = op.z = 0;
        for(int i=0;i<n;i++) 
            scanf("%lf %lf %lf", &p[i].x, &p[i].y, &p[i].z);
        
        solve();
    
    return 0;

---

POJ2069 Super Star

这一题三分做法会T，只能用模拟退火才能过。
注意初始点选择。

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps = 1e-5;
struct Point
    double x, y, z;
p[35], op;
int n;

inline double dist(Point &a, Point &b) 
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));

void solve() 
    double ans, delta = 100.0;
    double maxDis, tempDis;
    while(delta>eps)
        int id = 0;
        maxDis = dist(op, p[id]);
        for(int i=1;i<n;i++)
            tempDis=dist(op,p[i]);
            if(tempDis>maxDis)
                maxDis = tempDis;
                id = i;
            
        
        ans = maxDis;
        op.x += (p[id].x-op.x)/maxDis*delta;
        op.y += (p[id].y-op.y)/maxDis*delta;
        op.z += (p[id].z-op.z)/maxDis*delta;
        delta *= 0.98;
    
    printf("%.5lf\n", ans);


int main() 
    while(scanf("%d", &n)!=EOF && n) 
        op.x = op.y = op.z = 0;
        for(int i=0;i<n;i++) 
            scanf("%lf %lf %lf", &p[i].x, &p[i].y, &p[i].z);
            op.x += p[i].x;
            op.y += p[i].y;
            op.z += p[i].z;
        
        op.x /= n; op.y /= n; op.z /= n;
        solve();
    
    return 0;

---

HDU3007 HDU3932 类似。
注意HDU3932 n==1采用模拟退火要特判。。。。