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最小覆盖圆 || 模拟退火均可。

之前模拟退火弄得不太清楚，今天好好学了下别人的= = 。。。发现我的退火那么水那么水 = =。。。

不过这个如果掌握不好步长的话，还是会死很惨。

时间差别很大，最小覆盖圆0ms，退火好几百ms。。。

最小覆盖圆：

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <string>#include <algorithm>#define MID(x,y) ( ( x + y ) >> 1 )#define L(x) ( x << 1 )#define R(x) ( x << 1 | 1 )#define BUG puts("here!!!")using namespace std;const double eps = 1e-6;const int MAX = 1010;struct point{double x,y;};point p[MAX];bool dy(double x,double y){return x > y + eps;}// x > y bool xy(double x,double y){return x < y - eps;}// x < y bool dyd(double x,double y){ return x > y - eps;}// x >= y bool xyd(double x,double y){return x < y + eps;} // x <= y bool dd(double x,double y) {return fabs( x - y ) < eps;}  // x == ydouble disp2p(point a,point b) //  a b 两点之间的距离 {return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );}point circumcenter(point a,point b,point c){ point ret; double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1*a1 + b1*b1)/2;   double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2*a2 + b2*b2)/2;   double d = a1 * b2 - a2 * b1;   ret.x = a.x + (c1*b2 - c2*b1)/d;   ret.y = a.y + (a1*c2 - a2*c1)/d;  return ret;}void min_cover_circle(point p[],int n,point &c,double &r){random_shuffle(p,p+n);// #include <algorithm>c = p[0]; r = 0;for(int i=1; i<n; i++)if( dy(disp2p(p[i],c),r) ){c = p[i]; r = 0;for(int k=0; k<i; k++)if( dy(disp2p(p[k],c),r) ){c.x = (p[i].x + p[k].x)/2;c.y = (p[i].y + p[k].y)/2;r = disp2p(p[k],c);for(int j=0; j<k; j++)if( dy(disp2p(p[j],c),r) ){// 求外接圆圆心，三点必不共线 c = circumcenter(p[i],p[k],p[j]);r = disp2p(p[i],c);}}}}int main(){int ncases,n;double x,y;while( ~scanf("%lf%lf%d",&x,&y,&n) ){for(int i=0; i<n; i++)scanf("%lf%lf",&p[i].x,&p[i].y);double r;point c;min_cover_circle(p,n,c,r);printf("(%.1lf,%.1lf).\n",c.x,c.y);printf("%.1lf\n",r);}return 0;}

模拟退火：

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <string>#include <algorithm>#include <time.h>#define MID(x,y) ( ( x + y ) >> 1 )#define L(x) ( x << 1 )#define R(x) ( x << 1 | 1 )#define BUG puts("here!!!")using namespace std;const int MAX = 1010;const double inf = 1e30;const double eps = 1e-8;const double pi = acos(-1.0);const int N = 10;const int L = 40;bool dy(double x,double y)    {    return x > y + eps;}    // x > y bool xy(double x,double y)    {    return x < y - eps;}    // x < y bool dyd(double x,double y)    {     return x > y - eps;}    // x >= y bool xyd(double x,double y)    {    return x < y + eps;}     // x <= y bool dd(double x,double y)     {    return fabs( x - y ) < eps;}  // x == ystruct point { double x,y;    point (double x,double y):x(x),y(y){}    point ():x(0),y(0){}};double disp2p(point a,point b) //  a b 两点之间的距离 {    return sqrt(( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ));}point p[MAX];point rp[MAX];double len[MAX];double min_dis(point a,point *p,int n){    double min = 0;    for(int i=0; i<n; i++)    {        double len = disp2p(a,p[i]);        if( dy(len,min) )            min = len;    }    return min;}bool check(point a,double x,double y){    return dyd(a.x,0.0) && dyd(a.y,0) && xyd(a.x,x) && xyd(a.y,y);}point Rand(double x,double y){point c;c.x = ( rand()%1000 + 1 ) / 1000.0 * x;c.y = ( rand()%1000 + 1 ) / 1000.0 * y;return c;}int main(){    int n,m,ncases;double x,y;srand(time(NULL));while( ~scanf("%lf%lf%d",&x,&y,&n) ){        for(int i=0; i<n; i++)            scanf("%lf%lf",&p[i].x,&p[i].y);        for(int i=0; i<N; i++)        {rp[i] = Rand(x,y);len[i] = min_dis(rp[i],p,n);}        point st;        double step = max(x,y)/sqrt((double)n);        while( step > 0.001 )        {            for(int k=0; k<N; k++)            {st = rp[k];                for(int i=0; i<L; i++)                {double ang = (rand()%1000+1)/1000.0*10*pi;double xx = st.x + step*cos(ang);double yy = st.y + step*sin(ang);                    point t = point(xx,yy);                    if( !check(t,x,y) ) continue;                    double dis = min_dis(t,p,n);                    if( xy(dis,len[k]) )                    {                        rp[k] = t;                        len[k] = dis;                    }                }}            step *= 0.8;        }        int ind = 0;        for(int i=1; i<N; i++)        if( len[i] < len[ind] )        ind = i;printf("(%.1lf,%.1lf).\n",rp[ind].x,rp[ind].y);printf("%.1lf\n",len[ind]);    }return 0;}