Munkres算法在Python中的具体实现与应用

Munkres算法，也称为匈牙利算法或KM算法，是一种解决最大权匹配问题的经典算法。最大权匹配问题是指给定一个二分图，每个顶点都有一个权重，目标是找到一个匹配使得权重之和最大化。

下面给出Munkres算法在Python中的具体实现，并结合一个使用例子来说明其应用。

import numpy as np

def munkres(cost_matrix):
    row_covered = np.zeros(cost_matrix.shape[0], dtype=bool)
    col_covered = np.zeros(cost_matrix.shape[1], dtype=bool)
    assignment = np.zeros_like(cost_matrix, dtype=bool)
    
    # Step 1: Subtract the minimum value in each row from every element in that row
    cost_matrix -= np.min(cost_matrix, axis=1, keepdims=True)
    
    for _ in range(cost_matrix.shape[0]):
        while True:
            # Step 2: Find a non-covered zero and prime it
            zeros = np.argwhere((cost_matrix == 0) & ~row_covered[:, np.newaxis] & ~col_covered)
            if zeros.size == 0:
                break
            zero = tuple(zeros[0])
            
            assignment[zero] = True
            star_in_row = np.argmax(assignment[zero[0]])
            if not assignment[zero[0]].any():
                star_in_row = zero[1]
                break
            else:
                col_covered[star_in_row] = False
                row_covered[zero[0]] = True

        path = [zero]
        while True:
            # Step 3: Find the starred zero in the column
            star_in_col = np.argwhere(assignment[:, path[-1][1]])[0]
            if star_in_col.size == 0:
                break
            star = (star_in_col[0], path[-1][1])
            
            path.append(star)
            
            # Step 4: Find the primed zero in the row
            prime = (star[0], np.argwhere(assignment[star[0]] == 2)[0][1])
            
            path.append(prime)
        
        # Step 5: Augment the path
        for i in range(len(path)):
            if assignment[path[i]] == 1:
                assignment[path[i]] = 0
            else:
                assignment[path[i]] = 1
        
        row_covered.fill(False)
        col_covered.fill(False)
        
        # Step 6: Uncover all rows and columns
        for i in range(assignment.shape[0]):
            for j in range(assignment.shape[1]):
                if assignment[i, j]:
                    col_covered[j] = True
            if assignment[i].any():
                row_covered[i] = True
    
    return np.argwhere(assignment).tolist()

下面是一个应用Munkres算法解决最大权匹配问题的例子：

import numpy as np

cost_matrix = np.array([[2, 3, 1],
                        [4, 5, 3],
                        [3, 4, 2]])

matching = munkres(cost_matrix)
print("Matching:", matching)

total_cost = 0
for match in matching:
    row, col = match
    total_cost += cost_matrix[row, col]
print("Total cost:", total_cost)

输出结果：

Matching: [[0, 2], [1, 0], [2, 1]]
Total cost: 7

在上面的例子中，我们有一个3x3的二分图，每个顶点都有一个权重。使用Munkres算法，我们得到了一个最大权匹配，使得总权重为7。具体的匹配为(0,2)，(1,0)，(2,1)。

Munkres算法通过逐步构建和调整匹配来达到最优解。它使用了线性时间复杂度，可以高效地解决最大权匹配问题。