HDU - 5017 Ellipsoid(模拟退火法)

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Given a 3-dimension ellipsoid(椭球面)

your task is to find the minimal distance between the original point (0,0,0) and points on the ellipsoid. The distance between two points (x₁,y₁,z₁) and (x₂,y₂,z₂) is defined as 

1 0.04 0.01 0 0 0

1.0000000

思路：模拟退火法，学着网上写的

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int inf = 1e8;
const double eps = 1e-8;

const int dx[8] = {0,0,1,-1,1,-1,1,-1};
const int dy[8] = {1,-1,0,0,1,1,-1,-1};
double a, b, c, d, e, f;

double dis(double x, double y, double z) {
	return sqrt(x * x + y * y + z * z);
}

double calz(double x, double y) {
	double A = c;
	double B = d * y + e * x;
	double C = f * x * y + a * x * x + b * y * y - 1.0;
	double delta = B * B - 4.0 * A * C;

	if (delta < 0.0) return inf+10.0;
	delta = sqrt(delta);
	double z1 = (-B + delta) / (2.0 * A);
	double z2 = (-B - delta) / (2.0 * A);
	if (dis(x, y, z1) < dis(x, y, z2))
		return z1;
	return z2;
}

double solve() {
	double x = 0, y = 0, z = sqrt(1.0/c);
	double step = 1.0, rate = 0.99;
	while (step > eps) {
		for (int k = 0; k < 8; k++) {
			double nx = x + step * dx[k];
			double ny = y + step * dy[k];
			double nz = calz(nx, ny);

			if (nz >= inf) continue;
			if (dis(nx, ny, nz) < dis(x, y, z)) {
				x = nx;
				y = ny;
				z = nz;
			}
		}
		step *= rate;
	}
	return dis(x, y, z);
}

int main() {
	while (scanf("%lf%lf%lf%lf%lf%lf", &a, &b, &c, &d, &e, &f) != EOF) {
		printf("%.7lf\n", solve());
	}	
	return 0;
}

HDU - 5017 Ellipsoid(模拟退火法)

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原文地址：http://blog.csdn.net/u011345136/article/details/39779625