The Hungarian algorithm is a method for solving assignment problems. It is named after its inventor, George Dantzig, who published it in 1951. The algorithm is based on the idea of finding a minimum-cost matching between two sets of items.

The Hungarian algorithm can be used to solve a variety of problems, including:

The assignment problem: Given a set of workers and a set of jobs, find a way to assign each worker to a job so that the total cost is minimized.
The transportation problem: Given a set of suppliers and a set of customers, find a way to transport goods from the suppliers to the customers so that the total cost is minimized.
The minimum-cost flow problem: Given a network of nodes and edges, find a way to flow a certain amount of fluid from one node to another so that the total cost is minimized.
The Hungarian algorithm is a polynomial-time algorithm, meaning that it takes time that is proportional to the cube of the number of items in the problem. It is also a greedy algorithm, meaning that it always makes the choice that looks best at the moment, without considering the long-term consequences.

The Hungarian algorithm has been shown to be optimal in many cases, but it is not always the best choice for solving assignment problems. In some cases, other algorithms, such as the network simplex algorithm, can find a better solution.

Here is a sample implementation in python.
```python

import numpy as np

def min_zero_row(zero_mat, mark_zero):
	
	'''
	The function can be splitted into two steps:
	#1 The function is used to find the row which containing the fewest 0.
	#2 Select the zero number on the row, and then marked the element corresponding row and column as False
	'''

	#Find the row
	min_row = [99999, -1]

	for row_num in range(zero_mat.shape[0]): 
		if np.sum(zero_mat[row_num] == True) > 0 and min_row[0] > np.sum(zero_mat[row_num] == True):
			min_row = [np.sum(zero_mat[row_num] == True), row_num]

	# Marked the specific row and column as False
	zero_index = np.where(zero_mat[min_row[1]] == True)[0][0]
	mark_zero.append((min_row[1], zero_index))
	zero_mat[min_row[1], :] = False
	zero_mat[:, zero_index] = False

def mark_matrix(mat):

	'''
	Finding the returning possible solutions for LAP problem.
	'''

	#Transform the matrix to boolean matrix(0 = True, others = False)
	cur_mat = mat
	zero_bool_mat = (cur_mat == 0)
	zero_bool_mat_copy = zero_bool_mat.copy()

	#Recording possible answer positions by marked_zero
	marked_zero = []
	while (True in zero_bool_mat_copy):
		min_zero_row(zero_bool_mat_copy, marked_zero)
	
	#Recording the row and column positions seperately.
	marked_zero_row = []
	marked_zero_col = []
	for i in range(len(marked_zero)):
		marked_zero_row.append(marked_zero[i][0])
		marked_zero_col.append(marked_zero[i][1])

	#Step 2-2-1
	non_marked_row = list(set(range(cur_mat.shape[0])) - set(marked_zero_row))
	
	marked_cols = []
	check_switch = True
	while check_switch:
		check_switch = False
		for i in range(len(non_marked_row)):
			row_array = zero_bool_mat[non_marked_row[i], :]
			for j in range(row_array.shape[0]):
				#Step 2-2-2
				if row_array[j] == True and j not in marked_cols:
					#Step 2-2-3
					marked_cols.append(j)
					check_switch = True

		for row_num, col_num in marked_zero:
			#Step 2-2-4
			if row_num not in non_marked_row and col_num in marked_cols:
				#Step 2-2-5
				non_marked_row.append(row_num)
				check_switch = True
	#Step 2-2-6
	marked_rows = list(set(range(mat.shape[0])) - set(non_marked_row))

	return(marked_zero, marked_rows, marked_cols)

def adjust_matrix(mat, cover_rows, cover_cols):
	cur_mat = mat
	non_zero_element = []

	#Step 4-1
	for row in range(len(cur_mat)):
		if row not in cover_rows:
			for i in range(len(cur_mat[row])):
				if i not in cover_cols:
					non_zero_element.append(cur_mat[row][i])
	min_num = min(non_zero_element)

	#Step 4-2
	for row in range(len(cur_mat)):
		if row not in cover_rows:
			for i in range(len(cur_mat[row])):
				if i not in cover_cols:
					cur_mat[row, i] = cur_mat[row, i] - min_num
	#Step 4-3
	for row in range(len(cover_rows)):  
		for col in range(len(cover_cols)):
			cur_mat[cover_rows[row], cover_cols[col]] = cur_mat[cover_rows[row], cover_cols[col]] + min_num
	return cur_mat

def hungarian_algorithm(mat): 
	dim = mat.shape[0]
	cur_mat = mat

	#Step 1 - Every column and every row subtract its internal minimum
	for row_num in range(mat.shape[0]): 
		cur_mat[row_num] = cur_mat[row_num] - np.min(cur_mat[row_num])
	
	for col_num in range(mat.shape[1]): 
		cur_mat[:,col_num] = cur_mat[:,col_num] - np.min(cur_mat[:,col_num])
	zero_count = 0
	while zero_count < dim:
		#Step 2 & 3
		ans_pos, marked_rows, marked_cols = mark_matrix(cur_mat)
		zero_count = len(marked_rows) + len(marked_cols)

		if zero_count < dim:
			cur_mat = adjust_matrix(cur_mat, marked_rows, marked_cols)

	return ans_pos

def ans_calculation(mat, pos):
	total = 0
	ans_mat = np.zeros((mat.shape[0], mat.shape[1]))
	for i in range(len(pos)):
		total += mat[pos[i][0], pos[i][1]]
		ans_mat[pos[i][0], pos[i][1]] = mat[pos[i][0], pos[i][1]]
	return total, ans_mat

def main():

	'''Hungarian Algorithm: 
	Finding the minimum value in linear assignment problem.
	Therefore, we can find the minimum value set in net matrix 
	by using Hungarian Algorithm. In other words, the maximum value
	and elements set in cost matrix are available.'''

	#The matrix who you want to find the minimum sum
	cost_matrix = np.array([[7, 6, 2, 9, 2],
				[6, 2, 1, 3, 9],
				[5, 6, 8, 9, 5],
				[6, 8, 5, 8, 6],
				[9, 5, 6, 4, 7]])
	ans_pos = hungarian_algorithm(cost_matrix.copy())#Get the element position.
	ans, ans_mat = ans_calculation(cost_matrix, ans_pos)#Get the minimum or maximum value and corresponding matrix.

	#Show the result
	print(f"Linear Assignment problem result: {ans:.0f}\n{ans_mat}")

	#If you want to find the maximum value, using the code as follows: 
	#Using maximum value in the cost_matrix and cost_matrix to get net_matrix
	profit_matrix = np.array([[7, 6, 2, 9, 2],
				[6, 2, 1, 3, 9],
				[5, 6, 8, 9, 5],
				[6, 8, 5, 8, 6],
				[9, 5, 6, 4, 7]])
	max_value = np.max(profit_matrix)
	cost_matrix = max_value - profit_matrix
	ans_pos = hungarian_algorithm(cost_matrix.copy())#Get the element position.
	ans, ans_mat = ans_calculation(profit_matrix, ans_pos)#Get the minimum or maximum value and corresponding matrix.
	#Show the result
	print(f"Linear Assignment problem result: {ans:.0f}\n{ans_mat}")

if __name__ == '__main__':
	main()
```### Instruction:
 is there a python library that I can use instead of this code?### Response:
 Yes, the `scipy` library implements the Hungarian algorithm in its `optimize` module: `scipy.optimize.linear_sum_assignment`. Should I integrate it in the previous code?
헝가리 알고리즘은 과제 문제를 푸는 방법입니다. 1951년에 이 알고리즘을 발표한 발명가 조지 단치그의 이름을 따서 명명되었습니다. 이 알고리즘은 두 세트의 항목 사이에서 최소 비용으로 일치하는 항목을 찾는 아이디어를 기반으로 합니다.

헝가리 알고리즘은 다음과 같은 다양한 문제를 해결하는 데 사용할 수 있습니다:

할당 문제: 작업자 집합과 작업 집합이 주어졌을 때, 총 비용을 최소화하도록 각 작업자를 작업에 할당하는 방법을 찾습니다.
운송 문제: 공급업체 집합과 고객 집합이 주어졌을 때, 총 비용이 최소화되도록 공급업체에서 고객으로 상품을 운송하는 방법을 찾습니다.
최소 비용 흐름 문제: 노드와 에지로 구성된 네트워크가 주어졌을 때, 총 비용을 최소화할 수 있도록 한 노드에서 다른 노드로 일정량의 유체를 흘려보내는 방법을 찾습니다.
헝가리 알고리즘은 다항식 시간 알고리즘으로, 문제에 포함된 항목 수의 세제곱에 비례하는 시간이 걸립니다. 또한 욕심 알고리즘으로, 장기적인 결과를 고려하지 않고 항상 그 순간에 가장 좋아 보이는 선택을 한다는 의미입니다.

헝가리 알고리즘은 많은 경우에 최적이라고 밝혀졌지만 과제 문제를 푸는 데 항상 최선의 선택은 아닙니다. 경우에 따라 네트워크 심플렉스 알고리즘과 같은 다른 알고리즘이 더 나은 해결책을 찾을 수 있습니다.

다음은 파이썬으로 구현한 샘플입니다.
```python

import numpy as np

def min_zero_row(zero_mat, mark_zero):
	
	'''
	이 함수는 두 단계로 나눌 수 있습니다:
	#1 이 함수는 0이 가장 적게 포함된 행을 찾는 데 사용됩니다.
	#2 행에서 0을 선택한 다음 해당 행과 열에 해당하는 요소를 False로 표시합니다.
	'''

	#Find the row
	min_row = [99999, -1]

	for row_num in range(zero_mat.shape[0]): 
		if np.sum(zero_mat[row_num] == True) > 0 and min_row[0] > np.sum(zero_mat[row_num] == True):
			min_row = [np.sum(zero_mat[row_num] == True), row_num]

	# Marked the specific row and column as False
	zero_index = np.where(zero_mat[min_row[1]] == True)[0][0]
	mark_zero.append((min_row[1], zero_index))
	zero_mat[min_row[1], :] = False
	zero_mat[:, zero_index] = False

def mark_matrix(mat):

	'''
	Finding the returning possible solutions for LAP problem.
	'''

	#Transform the matrix to boolean matrix(0 = True, others = False)
	cur_mat = mat
	zero_bool_mat = (cur_mat == 0)
	zero_bool_mat_copy = zero_bool_mat.copy()

	#Recording possible answer positions by marked_zero
	marked_zero = []
	while (True in zero_bool_mat_copy):
		min_zero_row(zero_bool_mat_copy, marked_zero)
	
	#Recording the row and column positions seperately.
	marked_zero_row = []
	marked_zero_col = []
	for i in range(len(marked_zero)):
		marked_zero_row.append(marked_zero[i][0])
		marked_zero_col.append(marked_zero[i][1])

	#Step 2-2-1
	non_marked_row = list(set(range(cur_mat.shape[0])) - set(marked_zero_row))
	
	marked_cols = []
	check_switch = True
	while check_switch:
		check_switch = False
		for i in range(len(non_marked_row)):
			row_array = zero_bool_mat[non_marked_row[i], :]
			for j in range(row_array.shape[0]):
				#Step 2-2-2
				if row_array[j] == True and j not in marked_cols:
					#Step 2-2-3
					marked_cols.append(j)
					check_switch = True

		for row_num, col_num in marked_zero:
			#Step 2-2-4
			if row_num not in non_marked_row and col_num in marked_cols:
				#Step 2-2-5
				non_marked_row.append(row_num)
				check_switch = True
	#Step 2-2-6
	marked_rows = list(set(range(mat.shape[0])) - set(non_marked_row))

	return(marked_zero, marked_rows, marked_cols)

def adjust_matrix(mat, cover_rows, cover_cols):
	cur_mat = mat
	non_zero_element = []

	#Step 4-1
	for row in range(len(cur_mat)):
		if row not in cover_rows:
			for i in range(len(cur_mat[row])):
				if i not in cover_cols:
					non_zero_element.append(cur_mat[row][i])
	min_num = min(non_zero_element)

	#Step 4-2
	for row in range(len(cur_mat)):
		if row not in cover_rows:
			for i in range(len(cur_mat[row])):
				if i not in cover_cols:
					cur_mat[row, i] = cur_mat[row, i] - min_num
	#Step 4-3
	for row in range(len(cover_rows)):  
		for col in range(len(cover_cols)):
			cur_mat[cover_rows[row], cover_cols[col]] = cur_mat[cover_rows[row], cover_cols[col]] + min_num
	return cur_mat

def hungarian_algorithm(mat): 
	dim = mat.shape[0]
	cur_mat = mat

	#Step 1 - Every column and every row subtract its internal minimum
	for row_num in range(mat.shape[0]): 
		cur_mat[row_num] = cur_mat[row_num] - np.min(cur_mat[row_num])
	
	for col_num in range(mat.shape[1]): 
		cur_mat[:,col_num] = cur_mat[:,col_num] - np.min(cur_mat[:,col_num])
	zero_count = 0
	while zero_count < dim:
		#Step 2 & 3
		ans_pos, marked_rows, marked_cols = mark_matrix(cur_mat)
		zero_count = len(marked_rows) + len(marked_cols)

		if zero_count < dim:
			cur_mat = adjust_matrix(cur_mat, marked_rows, marked_cols)

	return ans_pos

def ans_calculation(mat, pos):
	total = 0
	ans_mat = np.zeros((mat.shape[0], mat.shape[1]))
	for i in range(len(pos)):
		total += mat[pos[i][0], pos[i][1]]
		ans_mat[pos[i][0], pos[i][1]] = mat[pos[i][0], pos[i][1]]
	return total, ans_mat

def main():

	'''Hungarian Algorithm: 
	Finding the minimum value in linear assignment problem.
	Therefore, we can find the minimum value set in net matrix 
	by using Hungarian Algorithm. In other words, the maximum value
	and elements set in cost matrix are available.'''

	#The matrix who you want to find the minimum sum
	cost_matrix = np.array([[7, 6, 2, 9, 2],
				[6, 2, 1, 3, 9],
				[5, 6, 8, 9, 5],
				[6, 8, 5, 8, 6],
				[9, 5, 6, 4, 7]])
	ans_pos = hungarian_algorithm(cost_matrix.copy())#Get the element position.
	ans, ans_mat = ans_calculation(cost_matrix, ans_pos)#Get the minimum or maximum value and corresponding matrix.

	#Show the result
	print(f"Linear Assignment problem result: {ans:.0f}\n{ans_mat}")

	#If you want to find the maximum value, using the code as follows: 
	#Using maximum value in the cost_matrix and cost_matrix to get net_matrix
	profit_matrix = np.array([[7, 6, 2, 9, 2],
				[6, 2, 1, 3, 9],
				[5, 6, 8, 9, 5],
				[6, 8, 5, 8, 6],
				[9, 5, 6, 4, 7]])
	max_value = np.max(profit_matrix)
	cost_matrix = max_value - profit_matrix
	ans_pos = hungarian_algorithm(cost_matrix.copy())#Get the element position.
	ans, ans_mat = ans_calculation(profit_matrix, ans_pos)#Get the minimum or maximum value and corresponding matrix.
	#Show the result
	print(f"Linear Assignment problem result: {ans:.0f}\n{ans_mat}")

if __name__ == '__main__':
	main()
```### 지침:
 이 코드 대신 사용할 수 있는 파이썬 라이브러리가 있나요? ### 응답:
 예, `scipy` 라이브러리는 `optimize` 모듈에서 헝가리 알고리즘을 구현합니다: `scipy.optimize.linear_sum_assignment`입니다. 이전 코드에 통합해야 하나요?