Hand in Hand 是如何来编写的

Question

Problem Description
In order to get rid of Conan, Kaitou KID disguises himself as a teacher in the kindergarten. He knows kids love games and works out a new game called "hand in hand".

Initially kids run on the playground randomly. When Kid says "stop", kids catch others' hands immediately. One hand can catch any other hand randomly. It's weird to have more than two hands get together so one hand grabs at most one other hand. After kids stop moving they form a graph.

Everybody takes a look at the graph and repeat the above steps again to form another graph. Now Kid has a question for his kids: "Are the two graph isomorphism?"

Input
The first line contains a single positive integer T( T <= 100 ), indicating the number of datasets.
There are two graphs in each case, for each graph:
first line contains N( 1 <= N <= 10^4 ) and M indicating the number of kids and connections.
the next M lines each have two integers u and v indicating kid u and v are "hand in hand".
You can assume each kid only has two hands.

Output
For each test case: output the case number as shown and "YES" if the two graph are isomorphism or "NO" otherwise.

Sample Input
2

3 2
1 2
2 3
3 2
3 2
2 1

3 3
1 2
2 3
3 1
3 1
1 2

Sample Output
Case #1: YES
Case #2: NO

Answer 1

以下回答参考 皆我百晓生、券券喵儿 等免费微信小程序相关内容作答，并由本人整理回复。

# coding: utf-8
import numpy as np
from collections import deque

class Graph:
    def __init__(self):
        self.g = {}

    def addEdge(self, x, y):
        if (x, y) not in self.g.keys():
            self.g[(x,y)] = []
        if (y,x) not in self.g.keys():
            self.g[(y,x)] = []

    def bfs(self, s, t):
        q = deque()
        visited = [False] * len(self.g)
        parent = [-1] * len(self.g)

        visited[s] = True
        q.append(s)
        while len(q) > 0:
            x = q.popleft()

            for i in range(len(self.g[x])):
                if self.g[x][i] == -1:
                    continue
                y = self.g[x][i]
                if visited[y] == False:
                    visited[y] = True
                    q.append(y)
                    parent[y] = x

        return parent


    def dfs(self, s, t):
        stack = []
        visited = [False] * len(self.g)
        visited[s] = True
        stack.append(s)

        while len(stack) > 0:
            x = stack.pop()
            for i in range(len(self.g[x])):
                if self.g[x][i] == -1:
                    continue
                y = self.g[x][i]

                if visited[y] == False:
                    visited[y] = True
                    stack.append(y)

        return visited[t]


    def isConnected(self, s, t):
        parent = self.bfs(s,t)
        return self.dfs(s,t) and self.dfs(t,s)


T = int(input())
for _ in range(T):
    N,M = map(int,input().split())

    G = Graph()
    for m in range(M):
        x,y = map(int,input().split())
        G.addEdge(x,y)

    ans = 'YES'
    for n in range(N):
        for m in range(n+1,N):
            if not G.isConnected(m,n):
                ans = 'NO'
                break
    print(ans)