Python求解tsp问题（动态规划，简单易懂）

短命女 2022-05-19 04:15 612阅读 0赞

解题思路主要有两部分：

第一部分：i为当前节点（城市），S为还没有遍历的节点（城市集合），![\\varepsilon \\left ( i,S \\right )][varepsilon _left _ i_S _right]表示从第i个节点起，经历S集合中所有的点，到达终点的最短路径长度。因此有：![\\varepsilon \\left ( i,S \\right ) = min\\left ( d\\left ( i,j \\right )\_\{j\\in S\} + \\varepsilon \\left ( j,\{S\_\{/j\}\} \\right )\\right )][varepsilon _left _ i_S _right _ _ min_left _ d_left _ i_j _right _j_in S_ _ _varepsilon _left _ j_S_j_ _right _right]

第二部分，回溯，找到最优的路径，需要将S集合一一对应一个数字（类似于编码，一般用二进制），不如![S = \\left \\\{ 1,2,3 \\right \\\} \\rightarrow 2^1+2^2+2^3][S _ _left _ 1_2_3 _right _ _rightarrow 2_1_2_2_2_3]，然后比如从节点i等于0开始，未经历集合为![S = \\left \\\{ 1,2,3 \\right \\\}][S _ _left _ 1_2_3 _right]，而下一步最优的节点 j 等于2，那么![M\[i\]\[S\]=j][M_i_S_j]，回溯时只用从![M\[0\]\[S\]][M_0_S]向后推即可，例如![M\[0\]\[S\]=2\\rightarrow M\[2\]\[S\_\{/2\}\] = 3 \\rightarrow M\[3\]\[S\_\{/2,3\}\]][M_0_S_2_rightarrow M_2_S_2_ _ 3 _rightarrow M_3_S_2_3]，代码如下：

# tsp问题
    class Solution:
        def __init__(self,X,start_node):
            self.X = X #距离矩阵
            self.start_node = start_node #开始的节点
            self.array = [[0]*(2**len(self.X)) for i in range(len(self.X))] #记录处于x节点，未经历M个节点时，矩阵储存x的下一步是M中哪个节点
        def transfer(self,sets):
            su = 0
            for s in sets:
                su = su + 2**s # 二进制转换
            return su
        # tsp总接口
        def tsp(self):
            s = self.start_node
            num = len(self.X)
            cities = range(num) #形成节点的集合
            past_sets = [s] #已遍历节点集合
            cities.pop(cities.index(s)) #构建未经历节点的集合
            node = s #初始节点
            return self.solve(node,cities) #求解函数
        def solve(self,node,future_sets):
            # 迭代终止条件，表示没有了未遍历节点，直接连接当前节点和起点即可
            if len(future_sets) == 0:
                return self.X[node][self.start_node]
            d = 99999
            # node如果经过future_sets中节点，最后回到原点的距离
            distance = []
            # 遍历未经历的节点
            for i in range(len(future_sets)):
                s_i = future_sets[i]
                copy = future_sets[:]
                copy.pop(i) # 删除第i个节点，认为已经完成对其的访问
                distance.append(self.X[node][s_i] + self.solve(s_i,copy))
            # 动态规划递推方程，利用递归
            d = min(distance)
            # node需要连接的下一个节点
            next_one = future_sets[distance.index(d)]
            # 未遍历节点集合
            c = self.transfer(future_sets)
            # 回溯矩阵，（当前节点，未遍历节点集合）——>下一个节点
            self.array[node][c] = next_one
            return d
    D = [[-1,10,20,30,40,50],[12,-1,18,30,25,21],[23,19,-1,5,10,15],[34,32,4,-1,8,16],[45,27,11,10,-1,18],[56,22,16,20,12,-1]]
    S = Solution(D,0)
    print S.tsp()
    # 开始回溯
    M = S.array
    lists = range(len(S.X))
    start = S.start_node
    while len(lists) > 0:
        lists.pop(lists.index(start))
        m = S.transfer(lists)
        next_node = S.array[start][m]
        print start,"--->" ,next_node
        start = next_node

[varepsilon _left _ i_S _right]: https://private.codecogs.com/gif.latex?%5Cvarepsilon%20%5Cleft%20%28%20i%2CS%20%5Cright%20%29
[varepsilon _left _ i_S _right _ _ min_left _ d_left _ i_j _right _j_in S_ _ _varepsilon _left _ j_S_j_ _right _right]: https://private.codecogs.com/gif.latex?%5Cvarepsilon%20%5Cleft%20%28%20i%2CS%20%5Cright%20%29%20%3D%20min%5Cleft%20%28%20d%5Cleft%20%28%20i%2Cj%20%5Cright%20%29_%7Bj%5Cin%20S%7D%20&plus;%20%5Cvarepsilon%20%5Cleft%20%28%20j%2C%7BS_%7B/j%7D%7D%20%5Cright%20%29%5Cright%20%29
[S _ _left _ 1_2_3 _right _ _rightarrow 2_1_2_2_2_3]: https://private.codecogs.com/gif.latex?S%20%3D%20%5Cleft%20%5C%7B%201%2C2%2C3%20%5Cright%20%5C%7D%20%5Crightarrow%202%5E1&plus;2%5E2&plus;2%5E3
[S _ _left _ 1_2_3 _right]: https://private.codecogs.com/gif.latex?S%20%3D%20%5Cleft%20%5C%7B%201%2C2%2C3%20%5Cright%20%5C%7D
[M_i_S_j]: https://private.codecogs.com/gif.latex?M%5Bi%5D%5BS%5D%3Dj
[M_0_S]: https://private.codecogs.com/gif.latex?M%5B0%5D%5BS%5D
[M_0_S_2_rightarrow M_2_S_2_ _ 3 _rightarrow M_3_S_2_3]: https://private.codecogs.com/gif.latex?M%5B0%5D%5BS%5D%3D2%5Crightarrow%20M%5B2%5D%5BS_%7B/2%7D%5D%20%3D%203%20%5Crightarrow%20M%5B3%5D%5BS_%7B/2%2C3%7D%5D