最强大脑《数阵迷踪》python代码

最新推荐文章于 2024-08-18 16:53:58 发布

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最新推荐文章于 2024-08-18 16:53:58 发布

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文章标签： python 深度优先开发语言

本文链接：https://blog.youkuaiyun.com/qq_24484823/article/details/128872694

这篇博客介绍了如何使用Python的回溯算法解决《数阵迷踪》问题。代码注释详细说明了通过删除或保留边来确保最终图形由两个度为1的节点和n*n-2个度为2的节点组成，同时避免出现环路。运行结果显示算法成功达到预期目标。

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from collections import defaultdict
from math import gcd


class Vertex:
    def __init__(self, val, id):
        self.val = val
        self.id = id
        self.degree = 0

class Edge:
    def __init__(self, v1, v2):
        self.v1 = v1
        self.v2 = v2
    
    def set_score(self, graph):
        self.score = graph.vertex_set[self.v1].degree + graph.vertex_set[self.v2].degree

class Graph:
    def __init__(self, grid):
        self.n = len(grid)
        self.grid = grid
        self.vertex_set = {}
        self.edge_set = []
        self.edge_is_remove = []
        self.edge_visited = []
    
    def convert_edge_set(self):
        self.adj_map = defaultdict(list)
        for i in range(self.edge_num):
            if self.edge_is_remove[i] == 0:
                edge = self.edge_set[i]
                v1 = edge.v1
                v2 = edge.v2
                self.adj_map[v1].append(v2)
                self.adj_map[v2].append(v1)
    
    def query_linked(self, i, j, d=0):
        v1 = self.get_vertex_id(i, j)
        if d==0:
            v2 = self.get_vertex_id(i, j+1)
        else:
            v2 = self.get_vertex_id(i+1, j)
        return v2 in self.adj_map[v1]

    def get_vertex_id(self, i, j):
        id = i * self.n + j
        return id

    def build_graph(self):
        for i in range(self.n):
            for j in range(self.n):
                v1 = self.get_vertex_id(i, j)