题意:给了空间中的N个点及其坐标,求能覆盖所有点的最小球的半径
分析:这个题其实是属于计算几何的最小球覆盖问题,不过用模拟退火也可以解决,其思路是先对N个点求一下平均数找到一个还算"比较合适"的解,然后用O(n)的复杂度遍历该点和其余点的距离,找到最大的距离,最后用模拟退火的方式一点一点的逼近最远的点,最终找到最优解。
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 200;
struct Node {
double x, y, z;
}p[maxn];
Node ans;
double avex, avey, avez;
const double eps = 1e-8;
int N;
double dis(Node a, Node 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 SA() {
double T = 100;
double ret;
while (T>eps) {
int id = 1;
double max_dis = dis(ans, p[1]);
for (int i = 2; i <= N; i++) {
double tem = dis(ans, p[i]);
if (tem>max_dis) {
max_dis = tem;
id = i;
}
}
ret = max_dis;
ans.x += (p[id].x - ans.x) / max_dis*T;
ans.y += (p[id].y - ans.y) / max_dis*T;
ans.z += (p[id].z - ans.z) / max_dis*T;
T *= 0.98;
}
printf("%.5f\n", ret);
}
int main()
{
while (scanf("%d", &N) && N != 0) {
for (int i = 1; i <= N; i++) {
scanf("%lf%lf%lf", &p[i].x, &p[i].y, &p[i].z);
avex += p[i].x; avey += p[i].y; avez += p[i].z;
}
avex /= N; avey /= N; avez /= N;
ans.x = avex; ans.y = avey; ans.z = avez;
SA();
}
return 0;
}