uva216-Getting in Line(网络连线)

回溯-理解中。。。

寻找最短连线++++++（生成排列）

还要注意缩短时间的技巧----及时回溯，

代码经过努力有68ms优化到了24ms。。。

还是不知道人家那种0ms 的神代码怎么写的。先贴一下自己的代码吧！！

代码如下：

#include <cstdio>
#include <cmath>
#include <cstring>
int x[10], y[10], f[10], a[10], A[10];
double MIN, d[10], D[10];
void find_mindis(int n, int cur, double sum)
{
if(cur==n)
{
if(sum<MIN){ for(int i = 0; i < n; i++)
{
A[i] = a[i]; if(i) D[i] = d[i];
}
MIN = sum;
}
return;
}
if(sum>MIN)return;
for(int i = 1; i <= n; i++)
{
if(f[i])continue;
a[cur] = i;
f[i] = 1;
double temp = 0;
if(cur)
temp = d[cur] = sqrt((x[i]-x[a[cur-1]])*(x[i]-x[a[cur-1]])+(y[i]-y[a[cur-1]])*(y[i]-y[a[cur-1]]));
find_mindis(n, cur+1,sum+temp);
f[i] = 0;
}
}
int main ()
{
int n, Case = 0;
while(scanf("%d",&n)&&n)
{
MIN = 1000;
memset(f,0,sizeof(f));
for(int i = 1; i <= n; i++)
scanf("%d%d",&x[i],&y[i]);
printf("**********************************************************\nNetwork #%d\n", ++Case);
find_mindis(n, 0, 0);
for(int i = 1; i < n; i++) printf("Cable requirement to connect (%d,%d) to (%d,%d) is %.2lf feet.\n",x[A[i-1]],y[A[i-1]],x[A[i]],y[A[i]],D[i]+16);
printf("Number of feet of cable required is %.2lf.\n",MIN+16*n-16);
}
return 0;
}