poj2420

模拟退火算法，其实不如说随机化变步长的贪心，向四周搜索，如果碰到更小的则把它作为更优解。

本题代码中费马点部分可作为模版使用。

#include <iostream>#include <stdio.h>#include <math.h>#define MAX_NUM 128struct point{double x;double y;};double distance(point p1, point p2){return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));}//多边形费马点，点集pt，大小n，传入ptres作为费马点这一点，返回值是所有点到费马点的距离double fermat_point(point pt [], int n, point & ptres){point u, v;double step = 0.0, curlen, explen, minlen;int i, j, k, idx;bool flag;u.x = u.y = v.x = v.y = 0.0;for (i = 0; i < n; ++i){step += fabs(pt[i].x) + fabs(pt[i].y);u.x += pt[i].x;u.y += pt[i].y;}u.x /= n;u.y /= n;flag = 0;while (step > 1e-10){for (k = 0; k < 10; step /= 2, ++k)for (i = -1; i <= 1; ++i)for (j = -1; j <= 1; ++j){v.x = u.x + step*i;v.y = u.y + step*j;curlen = explen = 0.0;for (idx = 0; idx < n; ++idx){curlen += distance(u, pt[idx]);explen += distance(v, pt[idx]);}if (curlen > explen){u = v;minlen = explen;flag = 1;}}}ptres = u;return flag ? minlen : curlen;}int main(){//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);int n, i;double res;point pt[MAX_NUM], ptres;while (std::cin>>n){for (i = 0; i < n; ++i){std::cin>>pt[i].x>>pt[i].y;}res = fermat_point(pt, n, ptres);//std::cout << ptres.x << ' ' << ptres.y << std::endl;if (res - (int) (res) > 0.5)printf("%d\n", (int) (res + 1));elseprintf("%d\n", (int) (res));}}