看到数据范围那么小，一眼状压( ext{DP})。
设(dp[i][s])表示从(i)出发，走过的点的集合为(s)的最小距离。
不难推出转移方程（(dis(i,j))为(i)点到(j)点的距离）：
[dp[i][s] = min(dp[i][s], dp[j][s - (1 << (i - 1))] + dis(i,j))
]
边界：
[dp[i][1 << (i - 1)] = 0
]
代码实现：
#include <bits/stdc++.h>
#define itn int
#define gI gi
using namespace std;
inline int gi()
{
int f = 1, x = 0; char c = getchar();
while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return f * x;
}
int n;
double x[20], y[20], dis[20][20], ans = 1000000000.0, dp[20][1 << 16];
inline double getjuli(double x, double y, double xx, double yy)
{
return sqrt((x - xx) * (x - xx) + (y - yy) * (y - yy));
}
int main()
{
//freopen(".in", "r", stdin);
//freopen(".out", "w", stdout);
n = gi();
for (int i = 1; i <= n; i+=1) scanf("%lf %lf", &x[i], &y[i]);
for (int i = 1; i <= n; i+=1)
for (int j = 1; j <= n; j+=1)
dis[i][j] = getjuli(x[i], y[i], x[j], y[j]);//预处理出两点间的距离
memset(dp, 127, sizeof(dp));
for (int s = 1; s <= (1 << n) - 1; s+=1)
{
for (int i = 1; i <= n; i+=1)
{
if (!(s & (1 << (i - 1)))) continue;
if (s == (1 << (i - 1))) {dp[i][s] = 0; continue;}//边界
for (int j = 1; j <= n; j+=1)
{
if ((!s & (1 << (j - 1))) || i == j) continue;
dp[i][s] = min(dp[i][s], dp[j][s - (1 << (i - 1))] + dis[i][j]);//转移
}
}
}
for (int i = 1; i <= n; i+=1)
{
double ss = dp[i][(1 << n) - 1] + getjuli(0, 0, x[i], y[i]);//注意加上到点(0,0)的距离
if (ss < ans) ans = ss;
}
printf("%.2lf
", ans);
return 0;
}