/*提醒推荐：五星刘汝佳《算法竞赛入门经典》，集合上的动态规划---最优配对问题题意：空间里有n个点P0,P1,...,Pn-1，你的任务是把它们配成n/2对（n是偶数），使得每个点恰好在一个点对中。所有点对中两点的距离之和应尽量小。状态：d(i,S)表示把前i个点中，位于集合S中的元素两两配对的最小距离和状态转移方程为：d(i,S)=min{|PiPj|+d(i-1,S-{i}-{j}}书上的解法有些问题，正解见方法一方法二：状态可以进行压缩，i的值其实隐藏在S中，S中最高位为1的即为i，所以需要一次查找,从n-1到0进行一次历编即可，整个运算下来，平均查找次数仅为2。而且方法二比方法一情况简单很多，也比较容易理解。方法三：这道题用递归实现更好一些，因为只需要判断n为偶数的情况，这就是递归运算的好处，而非递归则需要全部都进行一次运算。技巧：①处使用有个技巧，传递引用而不是下标，书写会方便很多。*///方法一：正解。。。#include <cstdio>#include <cstring>#include <cmath>const int nMax=21;const double INF=1e10;int n;struct Node{int x,y,z;}node[nMax];double d[nMax][1<<nMax];void init(){scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d %d %d",&node[i].x,&node[i].y,&node[i].z);}double min(double a,double b){return a<b?a:b;}double dis(Node &a,Node &b)//①{return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));}void solve(){for(int i=0;i<n;i++){for(int s=0;s<(1<<(i+1));s++){if(s==0) d[i][s]=0;else d[i][s]=INF;if((s & (1<<i))){for(int j=i-1;j>=0;j--)if((s & (1<<j)))d[i][s]=min(d[i][s],dis(node[i],node[j])+d[i-1][s^(1<<i)^(1<<j)]);}else if(i!=0){d[i][s]=d[i-1][s];}}}}int main(){freopen("f://data.in","r",stdin);init();solve();printf("%.3lf\n",d[n-1][(1<<n)-1]);return 0;}//方法二：推荐。。。//#define TEST#include <cstdio>#include <cstring>#include <cmath>const int nMax=21;const double INF=1e10;int n,S;struct Node{int x,y,z;}node[nMax];double d[1<<nMax];void init(){scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d %d %d",&node[i].x,&node[i].y,&node[i].z);S=1<<n;for(int i=1;i<S;i++)d[i]=-1;d[0]=0;}double min(double a,double b){return a<b?a:b;}double dis(Node &a,Node &b){return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));}double dp(int p){if(d[p]!=-1) return d[p];d[p]=INF;int i,j;for(i=n-1;i>=0;i--)if(p & (1<<i))break;for(j=i-1;j>=0;j--)if(p & (1<<j))d[p]=min(d[p],dis(node[i],node[j])+dp(p^(1<<i)^(1<<j)));#ifdef TESTprintf("%d %d\n",p,d[p]);#endifreturn d[p];}int main(){freopen("f://data.in","r",stdin);init();printf("%.3lf\n",dp(S-1));return 0;}//方法三：递归实现#include <cstdio>#include <cstring>#include <cmath>const int nMax=21;const double INF=1e10;int n,S;struct Node{int x,y,z;}node[nMax];double d[1<<nMax];void init(){scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d %d %d",&node[i].x,&node[i].y,&node[i].z);S=1<<n;d[0]=0;}double min(double a,double b){return a<b?a:b;}double dis(Node &a,Node &b){return sqrt((double)(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z));}void solve(){for(int s=1;s<S;s++){int i,j;d[s]=INF;for(i=n-1;i>=0;i--)if(s & 1<<i)break;for(j=i-1;j>=0;j--)if(s & 1<<j)d[s]=min(d[s],dis(node[i],node[j])+d[s^(1<<i)^(1<<j)]);}}int main(){freopen("f://data.in","r",stdin);init();solve();printf("%.3lf\n",d[S-1]);return 0;}
