UVa 216 Getting in Line

#include<cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
struct Points
{
double x,y,dis;
}a[10],res[10];//a用来输入，res保存结果
double getdis(Points &a, Points &b)//计算距离
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))+16;
}
int main()
{
int n,k=1;
while(~scanf("%d",&n) && n)
{
printf("**********************************************************\n");
printf("Network #%d\n",k++);
int i,data[10];
double Min=2147483645,resd[10];
for(i=0;i<n;i++)
{
scanf("%lf%lf",&a[i].x,&a[i].y);
data[i]=i;        //记录每个点的索引
}
do{
double sum=0,tmpdis[10];
for(i=0;i<n-1;i++)
{
tmpdis[i]=getdis(a[data[i]],a[data[i+1]]);
sum+=tmpdis[i];
}
if(sum<Min)   // 如果找到更小的距离，保存点坐标与点间距
{
for(int j=0;j<n-1;j++)
{
resd[j]=tmpdis[j];
res[j]=a[data[j]];
}
res[n-1]=a[data[n-1]];
Min=sum;
}
}while (next_permutation(data,data+n));//下一个排列
for(i=0;i<n-1;i++)
printf("Cable requirement to connect (%.lf,%.lf) to (%.lf,%.lf) is %.2lf feet.\n",res[i].x,res[i].y,res[i+1].x,res[i+1].y,resd[i]);
printf("Number of feet of cable required is %.2lf.\n",Min);
}
return 0;
}