cogs 线型网络（状压dp）

/*
需要好大的空间.....
而且lowbit理解的不是很好
先放到博客里 以后慢慢研究
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define inf 999999999;
#define maxn 1050000
using namespace std;
int n,a[maxn];
double x[],y[],g[][],f[maxn][],ans=inf;
double Dis(double x1,double y1,double x2,double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int l(int x)
{
    return x&(-x);
}
int main()
{
    freopen("linec.in","r",stdin);
    freopen("linec.out","w",stdout);
    scanf("%d",&n);
    for(int i=;i<=n;i++)
      scanf("%lf%lf",&x[i],&y[i]);
    for(int i=;i<=n;i++)
      for(int j=;j<=n;j++)
        g[i][j]=Dis(x[i],y[i],x[j],y[j]);
    for(int i=;i<=n;i++)
      a[<<i]=i+;
    for(int i=;i<(<<n);i++)if(!a[i])
      for(int j=i;j;j-=l(j))//枚举状态i的每一个1
        {
          f[i][a[l(j)]]=inf;//更新此状态
          for(int k=i-l(j);k;k-=l(k))
            f[i][a[l(j)]]=min(f[i][a[l(j)]],f[i-l(j)][a[l(k)]]+g[a[l(j)]][a[l(k)]]);
        }
    for(int i=;i<=n;i++)
      ans=min(ans,f[(<<n)-][i]);
    printf("%.2f\n",ans);
    return ;
}