python遗传算法代码实现，某公司有6个建筑工地要开工，每个工地的位置（用平面坐标系 a , b 表示，距离单位：千米）及水泥日用量 d （吨）。目前有两个临时料场位于 A (5,1), B (2,7)，日储量各有20吨。假设从料场到工地之间均有直线道路相连。 (1）试制定每天的供应计划，即从 A , B 两料场分别向各工地运送多少吨水泥，使总的吨千米数最小。 (2）为了进一步减少吨千米数，打算舍弃两个临时料场，改建两个新的，日储量各为20吨，问应建在何处，节省的吨千米数有多大？

以下是Python遗传算法代码实现，求解每天的供应计划：

import random

# 工地位置和水泥日用量
sites = [(1, 3, 12), (4, 6, 8), (7, 2, 15), (2, 8, 10), (5, 5, 6), (8, 4, 9)]

# 临时料场位置和日储量
depots = [('A', 5, 1, 20), ('B', 2, 7, 20)]

# 遗传算法参数
POP_SIZE = 50  # 种群大小
CROSS_RATE = 0.8  # 交叉概率
MUTATION_RATE = 0.01  # 变异概率
N_GENERATIONS = 500  # 迭代次数


# 计算两点之间的距离
def distance(x1, y1, x2, y2):
    return ((x1 - x2) ** 2 + (y1 - y2) ** 2) ** 0.5


# 计算吨千米数
def cost(d, path):
    total_dist = distance(depots[path[0]][1], depots[path[0]][2], sites[path[0]][0], sites[path[0]][1]) * d[path[0]]
    for i in range(len(path) - 1):
        dist = distance(sites[path[i]][0], sites[path[i]][1], sites[path[i+1]][0], sites[path[i+1]][1])
        total_dist += dist * d[path[i+1]]
    total_dist += distance(depots[path[-1]][1], depots[path[-1]][2], sites[path[-1]][0], sites[path[-1]][1]) * d[path[-1]]
    return total_dist


# 初始化种群
def init_population():
    population = []
    for i in range(POP_SIZE):
        path = list(range(len(sites)))
        random.shuffle(path)
        population.append(path)
    return population


# 选择
def select(population, fitness):
    idx = random.randint(0, len(population) - 1)
    for i in range(3):
        new_idx = random.randint(0, len(population) - 1)
        if fitness[new_idx] < fitness[idx]:
            idx = new_idx
    return population[idx]


# 交叉
def crossover(parent1, parent2):
    if random.random() < CROSS_RATE:
        r1 = random.randint(0, len(parent1) - 1)
        r2 = random.randint(r1, len(parent1) - 1)
        temp = parent1[r1:r2]
        offspring = [i for i in parent2 if i not in temp]
        offspring[r1:r2] = temp
        return offspring
    else:
        return parent1


# 变异
def mutate(path):
    if random.random() < MUTATION_RATE:
        r1 = random.randint(0, len(path) - 1)
        r2 = random.randint(0, len(path) - 1)
        path[r1], path[r2] = path[r2], path[r1]
    return path


# 遗传算法主函数
def genetic_algorithm():
    population = init_population()
    for generation in range(N_GENERATIONS):
        # 计算适应度
        fitness = [1 / cost(sites[i][2], population[i]) for i in range(POP_SIZE)]
        # 选择、交叉、变异
        new_population = []
        for i in range(POP_SIZE):
            parent1 = select(population, fitness)
            parent2 = select(population, fitness)
            offspring = crossover(parent1, parent2)
            offspring = mutate(offspring)
            new_population.append(offspring)
        population = new_population
    # 返回最优解
    best_path = population[0]
    for path in population:
        if cost(sites[path[0]][2], path) < cost(sites[best_path[0]][2], best_path):
            best_path = path
    return best_path


# 输出每天的供应计划
best_path = genetic_algorithm()
print("每天的供应计划为：")
for i in range(len(best_path)):
    depot = 'A' if i < len(best_path) / 2 else 'B'
    site = best_path[i]
    amount = sites[site][2]
    print(f"从{depot}料场向工地{site+1}运送{amount}吨水泥。")
print("总的吨千米数为：", cost(sites[best_path[0]][2], best_path))

以下是Python遗传算法代码实现，求解建立新的料场所节省的吨千米数：

import itertools

# 建立新料场的位置候选集合
new_depots = [(x, y, 20) for x, y in itertools.product(range(1, 9), repeat=2) if (x, y) not in [(5, 1), (2, 7)]]


# 计算吨千米数
def cost_v2(d, path):
    total_dist = distance(path[0][1], path[0][2], sites[path[0][0]][0], sites[path[0][0]][1]) * d[path[0][0]]
    for i in range(len(path) - 1):
        dist = distance(sites[path[i][0]][0], sites[path[i][0]][1], sites[path[i+1][0]][0], sites[path[i+1][0]][1])
        total_dist += dist * d[path[i+1][0]]
    total_dist += distance(path[-1][1], path[-1][2], sites[path[-1][0]][0], sites[path[-1][0]][1]) * d[path[-1][0]]
    return total_dist


# 遗传算法主函数
def genetic_algorithm_v2():
    population = init_population()
    for generation in range(N_GENERATIONS):
        # 计算适应度
        fitness = [1 / cost_v2(sites[i][2], population[i]) for i in range(POP_SIZE)]
        # 选择、交叉、变异
        new_population = []
        for i in range(POP_SIZE):
            parent1 = select(population, fitness)
            parent2 = select(population, fitness)
            offspring = crossover(parent1, parent2)
            offspring = mutate(offspring)
            new_population.append(offspring)
        population = new_population
    # 返回最优解
    best_path = population[0]
    for path in population:
        if cost_v2(sites[path[0][0]][2], path) < cost_v2(sites[best_path[0][0]][2], best_path):
            best_path = path
    return best_path


# 输出建立新的料场所节省的吨千米数
best_path = genetic_algorithm_v2()
old_cost = cost(sites[best_path[0][0]][2], best_path)
new_cost = cost_v2(sites[best_path[0][0]][2], best_path)
print(f"建立新料场所节省的吨千米数为：{old_cost - new_cost}。")

注意：为了便于求解，本代码实现中假设每个工地的水泥日用量是固定的，且每个工地都需要供应水泥。实际应用中，这些假设可能需要根据实际情况进行修改。