题目：给你n台电脑所在的平面位置，求把他们连乘线型网络需要的最小的网线长度。

分析：搜索，枚举。

           因为数据规模很小，枚举所有电脑的全排列，每一个排列对应一种连线方式。

           枚举所有的连线方式，找到其中最小的，输出路径即可。

说明：开始以为是最短路或者最小生成树类似物(⊙_⊙)。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>

using namespace std;

typedef struct pnode
{
	int x,y;
}point;
point P[10];

double dist(point a, point b)
{
	return sqrt(0.0+(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))+16.0;
}

int  link[40320][10];
int  l_count;
int  used[10],save[10];
void dfs(int d, int n)
{
	if (d == n) {
		for (int i = 0 ; i < n ; ++ i)
			link[l_count][i] = save[i];
		l_count ++;
		return;
	}
	for (int i = 0 ; i < n ; ++ i)
		if (!used[i]) {
			used[i] = 1;
			save[d] = i;
			dfs(d+1, n);
			used[i] = 0;
		}
}

int main()
{
	int n,t = 1;
	while (~scanf("%d",&n) && n) {
		for (int i = 0 ; i < n ; ++ i)
			scanf("%d%d",&P[i].x,&P[i].y);
		
		for (int i = 0 ; i < n ; ++ i)
			used[i] = 0;
		l_count = 0;
		dfs(0, n);
		
		double min = 3000.0;
		int    spa = 0;
		for (int i = 0 ; i < l_count ; ++ i) {
			double sum = 0.0;
			for (int j = 1 ; j < n ; ++ j)
				sum += dist(P[link[i][j-1]], P[link[i][j]]);
			if (sum < min) {
				min = sum;
				spa = i;
			}
		}
		
		printf("**********************************************************\n");
		printf("Network #%d\n",t ++);
		for (int j = 1 ; j < n ; ++ j)
			printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n",			    P[link[spa][j-1]].x,P[link[spa][j-1]].y,P[link[spa][j]].x,P[link[spa][j]].y,					dist(P[link[spa][j-1]], P[link[spa][j]]));
		printf("Number of feet of cable required is %.2lf.\n",min);
	}
	return 0;
}

UVa 216 - Getting in Line