算法5·贪婪算法（旅行商问题）_贪心算法的算法matlab旅行商问题

解释:
该算法用于需要计算所有的解，并从中找到最短的那一个
与狄克斯特拉算法不同的是：这里只输入了狄克斯特拉算法里面的cost

# 输入旅行表信息
distance = {}
distance['ab'] = distance['ba'] = 10
distance['ac'] = distance['ca'] = 12
distance['ad'] = distance['da'] = 16
distance['bc'] = distance['cb'] = 12
distance['bd'] = distance['db'] = 10
distance['cd'] = distance['dc'] = 8


def search2(data, will_del, end, count):

    lowest_distance = 1000
    lowest_distance_name = None

    for i in distance.keys():
        if i[0] == data and i[-1] != will_del:  # 选择出发打头的路线并且剔除返回路线
            if distance[i] < lowest_distance:
                lowest_distance = distance[i]
                lowest_distance_name = i
    print(lowest_distance_name[-1])
    print("|")
    data = lowest_distance_name[-1]
    will_del = lowest_distance_name[0]
    count += 1
    if count > end:  # 循环结束标志，即回到出发点
        print('Have a good day')
    else:
        search2(data, will_del, end, count)


def main():
    data = input('请自定义起始出发点：')
    counts = input("请输入旅行地点个数：")
    print("最佳旅行路线如下：")
    print(data)
    print("|")
    count = eval(counts)  # 地点数
    end = count + (count - 1)
    search2(data, "N1", end, count)  # 输入N1,所有路线全是英文字母，1有利于首次剔除


if __name__ == '__main__':
    main()

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结果展示：

请自定义起始出发点：c
请输入旅行地点个数：4
最佳旅行路线如下：
c
|
d
|
b
|
a
|
c
|
Have a good day

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添加删除，使得再次循环次数减少
该方法可以在大量路线循环时减少运算时间，但是结果只能趋近于最优解（如该代码中的出发地b），这也是贪婪算法的弊端所在

# 输入旅行表信息
distance = {}
distance['ab'] = distance['ba'] = 10
distance['ac'] = distance['ca'] = 12
distance['ad'] = distance['da'] = 16
distance['bc'] = distance['cb'] = 12
distance['bd'] = distance['db'] = 10
distance['cd'] = distance['dc'] = 8


def search(data, will_del, end, count):
    lowest_distance = 1000
    lowest_distance_name = None

    for i in list(distance.keys()):
        if i[0] == data and i[-1] != will_del:  # 选择出发打头的路线并且剔除返回路线
            if distance[i] < lowest_distance:
                lowest_distance = distance[i]
                lowest_distance_name = i
                del distance[i]                 # 删除已过滤的路线，使得下次循环次数减少
                del distance[i[::-1]]

    print("     " + lowest_distance_name[-1])
    print("     |")
    data = lowest_distance_name[-1]
    will_del = lowest_distance_name[0]
    count += 1
    if count > end:  # 循环结束标志，即回到出发点
        print('回到出发点:')
        print('     |')
    else:
        search(data, will_del, end, count)


def main():
    data = input('请自定义起始出发点：')
    counts = input("请输入旅行地点个数：")
    print("最佳旅行路线：")
    print('     ' + data)
    print("     |")
    count = eval(counts)  # 地点数
    end = count + (count - 2)
    search(data, "N1", end, count)  # 输入N1,所有路线全是英文字母，1有利于首次剔除
    print("     " + data)
    print("     |")
    print("Have a good day")


if __name__ == '__main__':
    main()

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结果展示：

请自定义起始出发点：c
请输入旅行地点个数：4
最佳旅行路线：
     c
     |
     d
     |
     b
     |
     a
     |
回到出发点:
     |
     c
     |
Have a good day

Process finished with exit code 0

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