UVa 216 Getting in Line【枚举排列】

题意：给出n个点的坐标，（2<=n<=8）,现在要使得这n个点连通，问最小的距离的和

因为n很小，所以可以直接枚举这n个数的排列，算每一个排列的距离的和，

保留下距离和最小的那个排列就可以了（这个地方和带宽那题有点像，用memcpy将整个排列保留下来）

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include <cmath>
 #include<stack>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<algorithm>
 using namespace std;
 
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 
 typedef long long LL;
 const int INF = (<<)-;
 const int mod=;
 const int maxn=;
 
 int p[maxn],bestp[maxn];
 
 struct node{
     int x,y;
 } a[maxn];
 
 double dis(int x1,int y1,int x2,int y2){
      return sqrt((double)(x2 - x1) * (x2 - x1) +(double) (y2 - y1) * (y2 - y1));
  }
 
 int main(){
   // freopen("in.txt","r",stdin);
 //    freopen("out.txt","w",stdout);
     int  n;
     double ans,minn=;
     int kase=;
     while(scanf("%d",&n)!=EOF&&n){
         for(int i=;i<n;i++) {
             cin>>a[i].x>>a[i].y;
             if(i!=) minn+=dis(a[i].x,a[i].y,a[i-].x,a[i-].y);
         }
 
         for(int i=;i<n;i++) p[i]=i;
         memcpy(bestp,p,sizeof(p));
 
         while(next_permutation(p,p+n)){
             ans=;
             for(int i=;i<n;i++)
                     ans+=dis(a[p[i]].x,a[p[i]].y,a[p[i-]].x,a[p[i-]].y);                                
 
         //    printf("ans=%lf\n",ans);
 
                 if(ans<minn){
                     minn=ans;
                     memcpy(bestp,p,sizeof(p));
                 }
         }
 
         printf("**********************************************************\n");
         printf("Network #%d\n",++kase);
         for(int i = ; i < n; ++i)
             printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n",(int)a[bestp[i-]].x,
                     (int)a[bestp[i-]].y,(int)a[bestp[i]].x,(int)a[bestp[i]].y,16.0 + dis(a[bestp[i-]].x,a[bestp[i-]].y,a[bestp[i]].x,a[bestp[i]].y));
         printf("Number of feet of cable required is %.2lf.\n",minn + (n - ) * 16.0);
     }
     return ;
 }