Python3 回朔法解决作业分配问题 （剪枝优化）

本文是在上次的基础上做的优化，解决了穷举结果的尴尬

文章链接：Python3 回朔法解决作业分配问题（http://blog.csdn.net/liangxun0712/article/details/70598467）

废话不多说，直接上优化后的代码

class Worker:    max = 0  # 上界 通过贪心算法找出近似值    min = 0  # 下界 由每组的最小值组成    pt_nodes = []  # 存放可扩展的节点    input_file = ''  # 输入文件名    output_file = ''  # 输出文件名    matrix = []  # 存放数据矩阵  行为单个任务 每个工人 完成所要的时间    n = 0  # 数据矩阵的大小 n*n    min_cost = 10000    worker_mark = []  # 标记工人是否被分配了任务    #  初始化参数    def __init__(self, input_file, output_file):        self.input_file = input_file        self.output_file = output_file        self.read_data_from_file()        self.n = len(self.matrix)        self.get_low_limit()        self.get_up_limit()        print(self.matrix)        print(self.n)        print(self.max)        print(self.min)    #  从文件中读取数据 初始化数据矩阵    def read_data_from_file(self):        with open(self.input_file) as source:            for line in source:                data_cluster = line.split(',')                temp = []                for value in data_cluster:                    temp.append(int(value))                self.matrix.append(temp)    #  获取数据下界  最小值之和    def get_low_limit(self):        for i in range(self.n):            self.min += min(self.matrix[i])    #  获取数据上界  贪心算法    def get_up_limit(self):        #  初始化工人使用标记        worker_mark = []        for i in range(self.n):            worker_mark.append(0)        # 贪心算法 取得 近似最优解        for i in range(self.n):            temp = self.matrix[i]            min_value = 5000            index = 0            for k in range(self.n):                if worker_mark[k] == 0 and min_value > temp[k]:                    min_value = temp[k]                    index = k            worker_mark[index] = 1  # 标记工人是否被分配            self.max += min_value  # 累积上限值    #  初始化工人分配标记    def init_worker(self):        self.worker_mark = []        for i in range(self.n):            self.worker_mark.append(0)    #  剪枝优化后的回朔算法    def branch_limit(self, deep, total):        if deep == self.n:            # print('最短距离', total)            if self.min_cost > total:                self.min_cost = total            return        temp = self.matrix[deep]        for i in range(self.n):            if self.worker_mark[i] == 0:                if (temp[i] + total) > self.max:  # 超出上界的节点直接舍弃                    continue                else:                    self.worker_mark[i] = 1                    self.branch_limit(deep+1, total+temp[i])                    self.worker_mark[i] = 0input_file = 'input_assign05_01.dat'# input_file = 'input_assign05_02.dat'# input_file = 'input_assign05_03.dat'output_file = 'output_01.dat'# output_file = 'output_02.dat'# output_file = 'output_03.dat'worker = Worker(input_file, output_file)worker.init_worker()worker.branch_limit(0, 0)print(worker.min_cost)