旅行商问题的动态规划解法

一个售货员必须访问n个城市，这n个城市是一个完全图，售货员需要恰好访问所有城市的一次，并且回到最终的城市。
城市与城市之间有一个旅行费用，售货员希望旅行费用之和最少。
旅行商问题是np问题，一般可以使用回溯法或者动态规划解决。

class Solution:
    def __init__(self, X, start_node):
        self.X = X
        self.start_node = start_node
        self.array = [[0] * (2 ** (len(self.X) - 1)) for i in range(len(self.X))]

    def transfer(self, sets):
        su = 0
        for s in sets:
            su = su + 2 ** (s - 1)
        return su

    def tsp(self):
        s = self.start_node
        num = len(self.X)
        cities = list(range(num))
        cities.pop(cities.index(s))
        node = s
        return self.solve(node, cities)

    def solve(self, node, future_sets):
        if len(future_sets) == 0:
            return self.X[node][self.start_node]
        d = 9999999
        distance = []
        for i in range(len(future_sets)):
            s_i = future_sets[i]
            copy = future_sets[:]
            copy.pop(i)
            distance.append(self.X[node][s_i] + self.solve(s_i, copy))

        d = min(distance)
        next_one = future_sets[distance.index(d)]
        c = self.transfer(future_sets)
        self.array[node][c] = next_one
        return d


n = int(input())
m = int(input())
D = [[9999999 for j in range(n)] for i in range(n)]
for i in range(m):
    a, b, t = list(map(float, input().split()))
    a,b = int(a),int(b)
    D[a][b] = t
    D[b][a] = t

S = Solution(D, 0)
res = int(S.tsp())
if res >= 9999999:
    print(-1)
else:
    print(res)

参考：
漫画：什么是旅行商问题？
旅行商问题动态规划解法(python版)