python实现： 已知运输问题 Ai为产地，ai为产地Ai的产量（i=1,2,3）；Bj为销地，bj为销地Bj的销量（j=1,2,3,4）；每个方格右上角数字为产地Ai运输产品到销地Bj的单位运价cij （i=1,2,3, j=1,2,3,4）。 编程实现： （1）基于西北角法或最小元素法，设计实现求解该运输问题基本可行解的算法，确定该运输问题的一个基本可行解。 （2）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。

以下是Python实现，首先我们需要定义输入数据：

# 输入数据
ai = [10, 20, 30]
bj = [20, 25, 25, 15]
cij = [
    [2, 3, 1, 4],
    [5, 4, 8, 1],
    [3, 6, 2, 3]
]

其中，`ai`表示产地产量，`bj`表示销地销量，`cij`表示单位运价。

接下来使用西北角法求解基本可行解：

# 西北角法求解基本可行解
xi = [0] * len(ai)
xj = [0] * len(bj)
xij = [[0] * len(bj) for _ in range(len(ai))]
i, j = 0, 0
while i < len(ai) and j < len(bj):
    if ai[i] < bj[j]:
        xij[i][j] = ai[i]
        xj[j] = xj[j] + ai[i]
        xi[i] = xi[i] + ai[i]
        bj[j] = bj[j] - ai[i]
        i = i + 1
    else:
        xij[i][j] = bj[j]
        xj[j] = xj[j] + bj[j]
        xi[i] = xi[i] + bj[j]
        ai[i] = ai[i] - bj[j]
        j = j + 1

print("基本可行解：")
for i in range(len(ai)):
    for j in range(len(bj)):
        print(xij[i][j], end="\t")
    print()
print("总运费：", sum(xij[i][j] * cij[i][j] for i in range(len(ai)) for j in range(len(bj))))

最后，使用位势法和闭回路法求解最优解：

# 位势法和闭回路法求解最优解
u = [0] * len(ai)
v = [0] * len(bj)
for _ in range(100):
    # 求解位势u和v
    for i in range(len(ai)):
        for j in range(len(bj)):
            if xij[i][j] > 0:
                if u[i] == 0 and v[j] == 0:
                    u[i] = cij[i][j]
                elif u[i] == 0:
                    u[i] = cij[i][j] - v[j]
                elif v[j] == 0:
                    v[j] = cij[i][j] - u[i]
    # 求解闭回路
    delta = [[cij[i][j] - u[i] - v[j] for j in range(len(bj))] for i in range(len(ai))]
    closed_path = []
    for i in range(len(ai)):
        for j in range(len(bj)):
            if delta[i][j] == 0 and [i, j] not in closed_path:
                path = [[i, j]]
                while True:
                    for k in range(len(ai)):
                        if [k, path[-1][1]] in path and [k, j] not in path:
                            path.append([k, j])
                            break
                    else:
                        for k in range(len(bj)):
                            if [path[-1][0], k] in path and [i, k] not in path:
                                path.append([i, k])
                                break
                        else:
                            break
                closed_path.append(path)

    # 求解最小闭回路
    min_closed_path = min(closed_path, key=lambda x: sum(xij[i][j] for i, j in x))

    # 更新xij
    delta_x = min(xij[i][j] for i, j in min_closed_path)
    for i, j in min_closed_path:
        xij[i][j] = xij[i][j] - delta_x
        if xij[i][j] == 0:
            u[i] = 0
            v[j] = 0

    # 检查是否为最优解
    if all(delta[i][j] >= 0 for i in range(len(ai)) for j in range(len(bj))):
        break

print("最优解：")
for i in range(len(ai)):
    for j in range(len(bj)):
        print(xij[i][j], end="\t")
    print()
print("总运费：", sum(xij[i][j] * cij[i][j] for i in range(len(ai)) for j in range(len(bj))))

完整代码如下：