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 = (1<<30)-1;
const int mod=1000000007;
const int maxn=100005;

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=0;
    int kase=0;
    while(scanf("%d",&n)!=EOF&&n){
        for(int i=0;i<n;i++) {
            cin>>a[i].x>>a[i].y;
            if(i!=0) minn+=dis(a[i].x,a[i].y,a[i-1].x,a[i-1].y);            
        }
        
        for(int i=0;i<n;i++) p[i]=i; 
        memcpy(bestp,p,sizeof(p));
        
        
        while(next_permutation(p,p+n)){
            ans=0;
            for(int i=1;i<n;i++) 
                    ans+=dis(a[p[i]].x,a[p[i]].y,a[p[i-1]].x,a[p[i-1]].y);                                

        //    printf("ans=%lf
",ans);
                
                if(ans<minn){
                    minn=ans;
                    memcpy(bestp,p,sizeof(p));
                }
        }
        
        printf("**********************************************************
");
        printf("Network #%d
",++kase);
        for(int i = 1; i < n; ++i)
            printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.
",(int)a[bestp[i-1]].x,
                    (int)a[bestp[i-1]].y,(int)a[bestp[i]].x,(int)a[bestp[i]].y,16.0 + dis(a[bestp[i-1]].x,a[bestp[i-1]].y,a[bestp[i]].x,a[bestp[i]].y));
        printf("Number of feet of cable required is %.2lf.
",minn + (n - 1) * 16.0);
    }
    return 0;
}