Uva216—— Getting in Line

题目的意思求n个点连起来的最短距离。n最大为8，直接枚举排列的所以情况，然后求最短的距离，最短距离等于连线上的每两个点的距离加上16的和.

点的排序可以多种情况，只要找到最短距离就行。枚举排列用next_permutation函数。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

int n, data[15], x[15], y[15], ans[15];
double mindis;

double get_dis(int x1, int y1, int x2, int y2)
{
return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)) + 16.0;
}

void get()
{
double temp = 0.0;
for(int i = 0; i < n - 1; i++)
temp += get_dis(x[data[i]], y[data[i]], x[data[i + 1]], y[data[i + 1]]);

if(temp <= mindis)
{
mindis = temp;
for(int i = 0; i < n; i++)
ans[i] = data[i];
}
}

int main()
{
//	freopen("1.txt", "r", stdin);
int i, t = 0;
while(cin >> n)
{
if(!n)
break;
mindis = 0xfffffff;
for(i = 0; i < n; i++)
{
cin >> x[i] >> y[i];
data[i] = i;
}
do
{
get();
}while(next_permutation(data, data + n));

printf("**********************************************************\n");
printf("Network #%d\n", ++t);
for(i = 0; i < n - 1; i++)
{
printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n", x[data[i]], y[data[i]],
x[data[i + 1]], y[data[i + 1]], get_dis(x[data[i]], y[data[i]], x[data[i + 1]], y[data[i + 1]]));
}
printf("Number of feet of cable required is %.2lf.\n", mindis);
}
return 0;
}