解题思路：可以去看一下有关于双调欧几里德旅行商问题的文章，我这里只是讲解一下为什么最后输出的是dp[n][n-1] + dis[n][n-1]，解决上次没有解决的问题

本来到最后面，要求的dp[n][n] ＝ min(dp[i][n] + dis[i][n])
因为 dp[i][n] = dp[i][n - 1] + dis[n][n-1]
所以 dp[n][n] = min(dp[i][n - 1] + dis[n][n-1] + dis[i][n])

又因为dp[n][n-1] = min(dp[i][n-1] + dis[i][n]),且dp[n][n-1]前面已经求到
所以dp[n][n] = dp[n][n - 1] + dis[n][n-1]

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 210
const double INF = 0x3f3f3f3f3f3f3f3f;

struct Point {
    int x, y;
}P[N];
double dp[N][N];
double dis[N][N];
int n;

double len(Point a, Point b) {
    return sqrt((a.x - b.x) * (a.x - b.x) * 1.0 + (a.y - b.y) * (a.y - b.y) * 1.0);
}

void init() {

    for (int i = 1; i <= n; i++)
        scanf("%d%d", &P[i].x, &P[i].y);

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            dis[i][j] = len(P[i], P[j]);
        }

}

void solve() {
    dp[2][1] = dp[1][2] = dis[1][2];

    for (int i = 3; i <= n; i++) {  
        dp[i][i - 1] = INF;
        for (int j = 1; j < i - 1; j++) {
            dp[i][j] = dp[i - 1][j] + dis[i - 1][i];
            dp[i][i - 1] = min(dp[i][i-1], dp[i-1][j] + dis[j][i]); 

        }
    }
    printf("%.2lf\n", dp[n][n-1] + dis[n][n - 1]);
}

int main() {
    while (scanf("%d", &n) != EOF) {
        init();
        solve(); 
    }
    return 0;
}