Blue Bridge Cup python group - shortest path

Blue Bridge Cup python group - shortest path

Topic description

This question is a fill-in-the-blank question. After calculating the result, use the output statement in the code to output the filled result.

As shown in the figure below, GG is an undirected graph, where the length of the blue edge is 11, the length of the orange edge is 22, and the length of the green edge is 33.

import os
import sys
'''floyd算法'''
r_list = [
   ["A", "E", 1],["A", "B", 2],["A", "C", 1],
    ["A", "D", 1],["A", "E", 1],["B", "G", 1],
    ["B", "J", 2],["C", "D", 3],["C", "G", 3],
    ["C", "F", 3],["D", "G", 2],["D", "H", 1],
    ["D", "I", 2],["E", "H", 1],["E", "I", 3],
    ["F", "J", 1],["F", "G", 1],["G", "K", 2],
    ["G", "I", 3],["H", "L", 2],["H", "I", 1],
    ["I", "M", 3],["J", "S", 2],["K", "N", 1],
    ["K", "L", 3],["L", "R", 1],["L", "M", 1],
    ["M", "N", 2],["M", "Q", 1],["M", "S", 1],
    ["N", "P", 1],["Q", "O", 1],["O", "R", 3],
    ["P", "O", 1],["R", "S", 1],
]

dp = [[float('INF') for node_start in range(50)] for node_end in range(50)]

for self in range(len(dp)):
    dp[self][self] = 0
for node in r_list:  # 初始化dp数组
    dp[ord(node[0])-ord('A')][ord(node[1])-ord('A')] = node[2]

for i in range(ord('Z')-ord('A')):
    for j in range(ord('Z')-ord('A')):
        for k in range(ord('Z')-ord('A')):
            dp[i][j] = min(dp[i][j], dp[i][k]+dp[k][j])

print(dp[0][ord('S')-ord('A')])

import os
import sys

fig = [
    ["A", "E", 1],["A", "B", 2],["A", "C", 1],
    ["A", "D", 1],["A", "E", 1],["B", "G", 1],
    ["B", "J", 2],["C", "D", 3],["C", "G", 3],
    ["C", "F", 3],["D", "G", 2],["D", "H", 1],
    ["D", "I", 2],["E", "H", 1],["E", "I", 3],
    ["F", "J", 1],["F", "G", 1],["G", "K", 2],
    ["G", "I", 3],["H", "L", 2],["H", "I", 1],
    ["I", "M", 3],["J", "S", 2],["K", "N", 1],
    ["K", "L", 3],["L", "R", 1],["L", "M", 1],
    ["M", "N", 2],["M", "Q", 1],["M", "S", 1],
    ["N", "P", 1],["Q", "O", 1],["O", "R", 3],
    ["P", "O", 1],["R", "S", 1],
]

#所欲路线的距离的集合
nums = []

#这个方法能够求出每一条路线
def get(num, element):

    for x , y in enumerate(fig): #利用循环将每一个点的下一个点都遍历完全
        if element[1] == y[0]:
            num += y[2]
            if y[1] == 'S':
                nums.append(num) #抵达终点S的时候可以将num的值进行存储
            else:
                get(num=num, element=y)  #没有到达终点的话可以继续进行

for i , j in enumerate(fig):
    if j[0] == 'A':
        get(num=j[2] , element=j)
print(min(nums))

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