python list assignment index_怎么修改 出现IndexError: list assignment index out of range

本文介绍了一个基于Python实现的弗洛伊德算法,用于解决带权图中的所有顶点对之间的最短路径问题。通过设置递归限制并使用三维数组存储节点间路径信息,文章详细展示了如何构建图、运行弗洛伊德算法以及输出最短路径及成本。

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import sys

sys.setrecursionlimit(100000000)

# 弗洛伊德算法

def floyd():

n = len(graph)

for k in range(n):

for i in range(n):

for j in range(n):

if graph[i][k] + graph[k][j] < graph[i][j]:

graph[i][j] = graph[i][k] + graph[k][j]

parents[i][j] = parents[k][j] # 更新父结点

# 打印路径

def print_path(i, j):

if i != j:

print_path(i, parents[i][j])

print(j, end='-->')

# Data [u, v, cost]

datas = [

[1, 6, 6],

[2, 6, 4],

[2, 7, 7],

[3, 7, 5],

[3, 8, 4],

[4, 8, 9],

[4, 9, 4],

[5, 9, 3],

[6, 10, 3],

[6, 11, 4],

[7, 11, 3],

[7, 12, 2],

[8, 12, 3],

[8, 13, 1],

[9, 13, 2],

[9, 14, 5],

[10, 15, 9],

[11, 15, 3],

[11, 16, 6],

[12, 16, 4],

[12, 17, 2],

[13, 17, 1],

[13, 18, 4],

[14, 18, 3],

[15, 19, 2],

[15, 20, 5],

[16, 20, 4],

[16, 21, 3],

[17, 21, 7],

[17, 22, 7],

[18, 22, 1],

[18, 23, 6],

]

n = 23

# 无穷大

inf = 9999999999

# 构图

graph = [[(lambda x: 0 if x[0] == x[1] else inf)([i, j]) for j in range(n)] for i in range(n)]

parents = [[i] * n for i in range(2000)] # 关键地方,i-->j 的父结点初始化都为i

for u, v, c in datas:

graph[u][v] = c

floyd()

print('Costs:')

for row in graph:

for e in row:

print('∞' if e == inf else e, end='\t')

print()

print('\nPath:')

for i in range(n):

for j in range(n):

print('Path({}-->{}): '.format(i, j), end='')

print_path(i, j)

print(' cost:', graph[i][j])

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