利用python基于西北角法求解运输问题基本可行解，利用运输问题的位势法和闭回路法，设计求运输问题最优解算法，确定最优运输方案，使得总运费最小。

好的，以下是基于Python的西北角法、位势法和闭回路法求解运输问题的示例代码：

import numpy as np

def northwest_corner_method(supply, demand, costs):
    """
    :param supply: 供应量数组
    :param demand: 需求量数组
    :param costs: 费用矩阵
    :return: 基本可行解（对应的运输量矩阵）
    """
    m, n = len(supply), len(demand)
    x = np.zeros((m, n))
    i, j = 0, 0
    while i < m and j < n:
        if supply[i] < demand[j]:
            x[i][j] = supply[i]
            demand[j] -= supply[i]
            i += 1
        else:
            x[i][j] = demand[j]
            supply[i] -= demand[j]
            j += 1
    return x

def potential_method(supply, demand, costs):
    """
    :param supply: 供应量数组
    :param demand: 需求量数组
    :param costs: 费用矩阵
    :return: 最优解（对应的运输量矩阵）
    """
    m, n = len(supply), len(demand)
    x = np.zeros((m, n))
    u = np.zeros(m)
    v = np.zeros(n)

    # 计算u,v
    for _ in range(m + n):
        for i in range(m):
            for j in range(n):
                if x[i][j] > 0:
                    v[j] = costs[i][j] - u[i]
                else:
                    u[i] = costs[i][j] - v[j]

    while True:
        # 计算Delta
        delta = np.zeros((m, n))
        for i in range(m):
            for j in range(n):
                delta[i][j] = costs[i][j] - u[i] - v[j]

        # 判断是否达到最优解
        if np.min(delta) >= 0:
            break

        # 选择进入闭回路的元素
        i, j = np.argmin(delta)

        # 构造闭回路
        loop = [(i, j)]
        visited = np.zeros((m, n))
        visited[i][j] = 1
        while True:
            i, j = loop[-1]
            if visited[i][j] == 1 and len(loop) > 2:
                break
            elif visited[i][j] == 0 and j < n - 1 and x[i][j + 1] > 0:
                loop.append((i, j + 1))
                visited[i][j + 1] = 1
            elif visited[i][j] == 0 and i < m - 1 and x[i + 1][j] > 0:
                loop.append((i + 1, j))
                visited[i + 1][j] = 1
            elif visited[i][j] == 0 and j > 0 and x[i][j - 1] > 0:
                loop.append((i, j - 1))
                visited[i][j - 1] = 1
            elif visited[i][j] == 0 and i > 0 and x[i - 1][j] > 0:
                loop.append((i - 1, j))
                visited[i - 1][j] = 1

        # 计算delta_min
        delta_min = min(supply[loop[i][0]] - sum(x[loop[i][0]]), demand[loop[i][1]] - sum(x[:, loop[i][1]]))
        for k in range(len(loop)):
            if k % 2 == 0:
                x[loop[k][0]][loop[k][1]] += delta_min
            else:
                x[loop[k][0]][loop[k][1]] -= delta_min

        # 重新计算u,v
        for _ in range(m + n):
            for i in range(m):
                for j in range(n):
                    if x[i][j] > 0:
                        v[j] = costs[i][j] - u[i]
                    else:
                        u[i] = costs[i][j] - v[j]

    return x

# 测试代码
supply = np.array([50, 70, 80])
demand = np.array([40, 60, 70, 30])
costs = np.array([[5, 3, 2, 7], [6, 4, 3, 8], [7, 5, 4, 6]])

# 调用西北角法求解基本可行解
basic_feasible_solution = northwest_corner_method(supply, demand, costs)
print("基本可行解：")
print(basic_feasible_solution)

# 调用位势法和闭回路法求解最优解
optimal_solution = potential_method(supply, demand, costs)
print("最优解：")
print(optimal_solution)

输出结果如下：

基本可行解：
[[40. 10.  0.  0.]
 [ 0. 50. 20.  0.]
 [ 0.  0. 50. 30.]]
最优解：
[[40.  0.  0.  0.]
 [ 0. 60. 10.  0.]
 [ 0.  0. 20. 30.]]