UVA216-Getting in Line

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;

int b[8], c[8];

struct node{
	int x, y;
}a[10];

double lenth(double x1, double y1, double x2, double y2){

	return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)) + 16; 
} 

int main(){
	int n, t = 0;
	double len, sum, min;	
	while (scanf("%d", &n) && n){
		min = 100000;	
		for(int i = 0; i < n; i++){	
			scanf("%d %d", &a[i].x, &a[i].y);	
			b[i] = i;	
		}	
		
		do {
			sum = 0;	
			for(int i = 0; i < n - 1; i++){	
				len = lenth(a[b[i]].x, a[b[i]].y, a[b[i + 1]].x, a[b[i + 1]].y);	
				sum += len;	
			}	
			if (sum < min){	
				min = sum;	
				for(int i = 0; i < n; i++)	
					c[i] = b[i];
			}		
		} while (next_permutation(b, b + n));	
		printf("**********************************************************\n");	
		printf("Network #%d\n", ++t);
		for(int i = 0; i < n - 1; i++){
			len = lenth(a[c[i]].x, a[c[i]].y, a[c[i + 1]].x, a[c[i + 1]].y);	
			printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n", a[c[i]].x, a[c[i]].y, a[c[i + 1]].x, a[c[i + 1]].y, len);		
		}
		printf("Number of feet of cable required is %.2lf.\n", min); 	
		
	}	
	return 0;
}