POJ 2420 模拟退火 解题报告

A Star not a Tree?

Description

Luke wants to upgrade his home computer network from 10mbs to 100mbs. His existing network uses 10base2 (coaxial) cables that allow you to connect any number of computers together in a linear arrangement. Luke is particulary proud that he solved a nasty NP-complete problem in order to minimize the total cable length.
Unfortunately, Luke cannot use his existing cabling. The 100mbs system uses 100baseT (twisted pair) cables. Each 100baseT cable connects only two devices: either two network cards or a network card and a hub. (A hub is an electronic device that interconnects several cables.) Luke has a choice: He can buy 2N-2 network cards and connect his N computers together by inserting one or more cards into each computer and connecting them all together. Or he can buy N network cards and a hub and connect each of his N computers to the hub. The first approach would require that Luke configure his operating system to forward network traffic. However, with the installation of Winux 2007.2, Luke discovered that network forwarding no longer worked. He couldn’t figure out how to re-enable forwarding, and he had never heard of Prim or Kruskal, so he settled on the second approach: N network cards and a hub.

Luke lives in a loft and so is prepared to run the cables and place the hub anywhere. But he won’t move his computers. He wants to minimize the total length of cable he must buy.

Input

The first line of input contains a positive integer N <= 100, the number of computers. N lines follow; each gives the (x,y) coordinates (in mm.) of a computer within the room. All coordinates are integers between 0 and 10,000.

Output

Output consists of one number, the total length of the cable segments, rounded to the nearest mm.

Sample Input

4
0 0
0 10000
10000 10000
10000 0

Sample Output

28284

【解题报告】
模拟退火求费马点是非常经典的问题了，如果觉得模拟退火这个名字比较高端难以理解的话们可以把它叫做“随机化变步长贪心”。

据说某位大神说过：爬山算法就是一只兔子看到一座山峰，然后跳来跳去最后跳上山顶，模拟退火就是一只喝醉的兔子，一开始乱跳，过一会酒醒了，然后再跳上山顶——黄学长

代码如下：

#include<cstdio>#include<cstdlib>#include<cmath>#include<ctime>using namespace std;int n;double xx,yy,ans,t;struct point{    double x,y;}p[105];double dis(double x,double y,point p){    return sqrt((x-p.x)*(x-p.x)+(y-p.y)*(y-p.y));}double getsum(double x,double y){    double tmp=0;    for(int i=1;i<=n;++i)        tmp+=dis(x,y,p[i]);    return tmp;}int main(){//  srand(time(0));    while(scanf("%d",&n)!=EOF)    {        xx=yy=0,ans=1e20,t=100000;        for(int i=1;i<=n;++i)        {            scanf("%lf%lf",&p[i].x,&p[i].y);            xx+=p[i].x,yy+=p[i].y;        }        xx/=n;yy/=n;        ans=getsum(xx,yy);        double tmp,x,y;        while(t>0.02)        {            x=y=0;            for(int i=1;i<=n;++i)            {                x+=(p[i].x-xx)/dis(xx,yy,p[i]);                y+=(p[i].y-yy)/dis(xx,yy,p[i]);            }            tmp=getsum(xx+x*t,yy+y*t);            if(tmp<ans)            {                ans=tmp;                xx+=x*t,yy+=y*t;            }            else if(log((tmp-ans)/t)<(rand()%10000)/10000.0)            {                ans=tmp;                xx+=x*t,yy+=y*t;            }                   t*=0.9;         }        printf("%.0lf\n",ans);    }    return 0;}

优化过随机函数后，0ms。。。

#include<cstdio>#include<cstdlib>#include<cmath>#include<ctime>using namespace std;int n;double xx,yy,ans,t;struct point{    double x,y;}p[105];inline int q_rand(){    static int seed=10007;     return seed=int(seed*48271LL%2147483647);}inline double dis(double x,double y,point p){    return sqrt((x-p.x)*(x-p.x)+(y-p.y)*(y-p.y));}inline double getsum(double x,double y){    double tmp=0;    for(int i=1;i<=n;++i)        tmp+=dis(x,y,p[i]);    return tmp;}int main(){    while(scanf("%d",&n)!=EOF)    {        xx=yy=0,ans=1e20,t=100000;        for(int i=1;i<=n;++i)        {            scanf("%lf%lf",&p[i].x,&p[i].y);            xx+=p[i].x,yy+=p[i].y;        }        xx/=n;yy/=n;        ans=getsum(xx,yy);        double tmp,x,y;        while(t>0.02)        {            x=y=0;            for(int i=1;i<=n;++i)            {                x+=(p[i].x-xx)/dis(xx,yy,p[i]);                y+=(p[i].y-yy)/dis(xx,yy,p[i]);            }            tmp=getsum(xx+x*t,yy+y*t);            if(tmp<ans)            {                ans=tmp;                xx+=x*t,yy+=y*t;            }            else if(log((tmp-ans)/t)<(q_rand()%10000)/10000.0)            {                ans=tmp;                xx+=x*t,yy+=y*t;            }                   t*=0.9;         }        printf("%.0lf\n",ans);    }    return 0;}

让我看到你们的双手