模拟退火 (poj 2420, poj 2069)

模拟退火基本知识
其伪代码如下：

Let s = s0For k = 0 through k_max (exclusive):  T := temperature(k / k_max)Pick a random neighbour, s_new := neighbour(s)If P(E(s), E(s_new), T) > random(0, 1), move to the new state:s := s_newOutput: the final state s

例子：

poj 2420

题意：
平面上给你n个点(xi,yi)，让你求一个点，到这n点的距离和最小。

限制：
1 <= n <= 100
0 <= xi,yi <= 1e4, 为整数

/*poj 2420  题意：  平面上给你n个点(xi,yi)，让你求一个点，到这n点的距离和最小。  限制：  1 <= n <= 100  0 <= xi,yi <= 1e4, 为整数  思路：  模拟退火  模拟退火基本知识：  其伪代码如下：  Let s = s0  For k = 0 through k_max (exclusive):T := temperature(k / k_max)  Pick a random neighbour, s_new := neighbour(s)  If P(E(s), E(s_new), T) > random(0, 1), move to the new state:  s := s_new  Output: the final state s*/#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>using namespace std;const double E = exp(1.0);struct Pt{double x,y;}p[105];double sqr(double x){return x*x;}double dis(Pt a, Pt b){return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));}double get_sum(Pt p0, int n){double ret = 0;for(int i = 0; i < n; ++i)ret += dis(p0, p[i]);   return ret;}int main(){srand(1123);int n;int limit = 10000; //(x,y)的范围while(scanf("%d", &n) != EOF){double x0 = 0, y0 = 0;for(int i = 0; i < n; ++i){scanf("%lf%lf", &p[i].x, &p[i].y);x0 += p[i].x;y0 += p[i].y;}x0 /= n;y0 /= n;double ans = get_sum((Pt){x0, y0}, n);double temp = 1e5; //初始温度, 根据题目修改while(temp > 0.02){ //0.02为温度的下限, 若温度temp达到下限, 则停止搜索double x = 0, y = 0;for(int i = 0; i < n; ++i){ //获取步长的规则根据题目而定x += (p[i].x - x0) / dis((Pt){x0, y0}, p[i]);y += (p[i].y - y0) / dis((Pt){x0, y0}, p[i]);}double tmp = get_sum((Pt){x0 + x * temp, y0 + y * temp}, n); //标函数E(x_new);if(tmp < ans){ans = tmp;x0 += x * temp;y0 += y * temp;} else if(pow(E, (ans - tmp) / temp) > (rand() % limit) / (double)limit){ans = tmp;x0 += x * temp;y0 += y * temp;}temp *= 0.9; //0.9为降温退火速率, (范围为0~1, 越大得到全局最优解的概率越高, 运行时间越长}printf("%.0f\n", ans);}return 0;}

poj 2069 Super Star

题意：
给定n个点(xi, yi, zi), 要求覆盖这些点的最小球半径。

限制：
4 <= n <= 30
0 <= xi, yi, zi <= 100

/*poj 2069 Super Star  题意：  给定n个点(xi, yi, zi), 要求覆盖这些点的最小球半径。  限制：  4 <= n <= 30  0 <= xi, yi, zi <= 100  思路：  模拟退火 */#include<iostream>#include<cstdio>#include<cmath>#include<cstdlib>using namespace std;const double E = exp(1.0);struct Pt{double x, y, z;}p[35];double sqr(double x){return x * x;}double dis(Pt a, Pt b){return sqrt(sqr(a.x - b.x) +sqr(a.y - b.y) +sqr(a.z - b.z));}double get_max_r(Pt p0, int n){double ret = 0;for(int i = 0; i < n; ++i)ret = max(ret, dis(p0, p[i]));return ret;}int main() {int n;int limit = 100;while(scanf("%d", &n) && n){double x0 = 0, y0 = 0, z0 = 0;for(int i = 0; i < n; ++i){scanf("%lf%lf%lf", &p[i].x, &p[i].y, &p[i].z);x0 += p[i].x;y0 += p[i].y;z0 += p[i].z;}x0 /= n;y0 /= n;z0 /= n;//cout<<x0<<' '<<y0<<' '<<z0<<endl;double ans = get_max_r((Pt){x0, y0, z0}, n);double temp = 1e5;while(temp > 1e-8){double x = 0, y = 0, z = 0;double max_r = 0;for(int i = 0; i < n; ++i){double r = dis((Pt){x0, y0, z0}, p[i]);if(r > max_r){x = (p[i].x - x0) / r;y = (p[i].y - y0) / r;z = (p[i].z - z0) / r;max_r = r;}}double tmp = get_max_r((Pt){x0 + x * temp,y0 + y * temp,z0 + z * temp}, n);if(tmp < ans){ans = tmp;x0 += x * temp;y0 += y * temp;z0 += z * temp;} else if(pow(E, (ans - tmp) / temp) > (rand() % limit) / (double)limit){ans = tmp;x0 += x * temp;y0 += y * temp;z0 += z * temp;}temp *= 0.998;}printf("%.5f\n", ans);}return 0;}