| 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. 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 Sample Output Source |
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题意：给你N个点，让你在平面上选择一个点，使得这个点到所有点的距离和最短。
解题思路：模拟退火！网上很多代码其实是爬山但是却说成了退火。当然爬山也能过！具体看代码！
#include<bits/stdc++.h>
using namespace std;
const int MAXN=500005;
const double eps=1e-8;
const double V=0.98;
typedef long long ll;
int N;
double X[MAXN];
double Y[MAXN];
double dx[8]={-1,-1,0,1,1,1,0,-1};
double dy[8]={0,1,1,1,0,-1,-1,-1};
double dis(double x,double y){
double ret=0;
for(int i=1;i<=N;i++)
ret+=sqrt((x-X[i])*(x-X[i])+(y-Y[i])*(y-Y[i]));
return ret;
}
double rand01()
{
double t1=(double)(rand()%1000)/(double)1000,
t2=(double)(rand()%1000)/(double)1000;
double r=t1+t2*0.001;
return r;
}
double tuihuo(){
double x=rand()%10000,y=rand()%10000;
double ans=dis(x,y);
double dt=sqrt(10000*10000*2);
while(dt>eps){
//八个方向尝试，实际上4个足够了。
for(int i=0;i<8;i++){
double nx=x+dx[i]*dt;
double ny=y+dy[i]*dt;
double d=dis(nx,ny);
//如果更优，必然转移
if(d<ans){
x=nx;
y=ny;
ans=d;
}
else{
//模拟退火核心操作，即使答案不是最优，也有一定概率转移
if(exp(ans-d)/dt>rand01()){
x=nx;
y=ny;
ans=d;
}
}
}
dt*=V;//缩短步长
}
return ans;
}
int main(){
scanf("%d",&N);
for(int i=1;i<=N;i++)
scanf("%lf%lf",&X[i],&Y[i]);
printf("%.0lf\n",tuihuo());
return 0;
}