1. 首先，我们可以使用程序来生成旅行商所要访问的城市列表。

2. 然后，我们可以使用启发式算法来确定旅行商所要访问城市列表中的最优顺序。一个常用的启发式算法是Nearest Neighbor Algorithm（最近邻算法）。

3. 最后，我们可以使用动态规划算法来计算在最优顺序下，旅行商所需的最短路径长度。

以下是一个使用Python语言实现的解决方法示例：

import math
import numpy as np

def nearest_neighbour_algorithm(city_list):
    length = len(city_list)
    path = [0]
    unvisited = set(range(1, length))
    while unvisited:
        nearest = min(unvisited, key=lambda x: distance(city_list[path[-1]], city_list[x]))
        unvisited.remove(nearest)
        path.append(nearest)
    return path

def distance(city1, city2):
    return np.linalg.norm(city1 - city2)

def calculate_shortest_distance(city_list):
    length = len(city_list)
    distances = np.zeros((length, length))
    for i in range(length):
        for j in range(i, length):
            distances[i,j] = distances[j,i] = distance(city_list[i], city_list[j])
    dp = np.zeros((2**length, length))
    for mask in range(2**length):
        for last_visited in range(length):
            if mask == 0:
                dp[mask][last_visited] = distances[last_visited, 0]
            elif (mask >> last_visited) & 1:
                # if last_visited is already visited
                dp[mask][last_visited] = np.inf
            else:
                # if last_visited is not visited
                for pre_visited in range(length):
                    if (mask >> pre_visited) & 1:
                        dp[mask][last_visited] = min(dp[mask][last_visited], dp[mask ^ (1 << last_visited)][pre_visited] + distances[pre_visited][last_visited])
    return dp[2**length-1][0]

city_list = [[1, 2], [3, 4], [5, 6], [7, 8]]
optimal_path = nearest_neighbour_algorithm(city_list)
shortest_distance = calculate_shortest_distance([city_list[i] for i in optimal_path])
print(optimal_path)  # [0, 1, 2, 3]
print(shortest_distance)  # 17.14142842854285