Question

我正在尝试解决一个问题，我应该创建一个字体，该字典的坐标标记为1，其邻居也是一（1）。这基本上是我需要创建的邻接列表为了在它上面实现BFS / DFS。

我使用setdefault（）方法以及使用get（）方法尝试过相同的操作。我之前已经解决了一些问题，但我无法记住它。

这是我为此编写的代码。

n,m=input().split()

n=int(n)
m=int(m)
#q=int(q)

"""User Input Done"""
square=[[0 for i in range(n)] for j in range(m)]

neighbour_dictionary={}

#neighbour_dictionary.setdefault(,[])

for i in range(n):
    for j in range(m):
        number=int(input("Enter a number"))
        square[i][j]=number

"""Get Initial Number of Nations(1)"""
nation_count=0
for i in range(0,n):
    for j in range(0,m):
        if square[i][j]==1:

            try:
                if square[i][j+1]==1 and (j+1)>0 and (i<n) and (j<m):
                    try:
                        neighbour_dictionary[(i,j)].setdefault([i,j],[]).append((i,j+1))
                    except KeyError:
                        neighbour_dictionary[(i,j)]=[(i,j+1)]

            except IndexError:
                pass
                #print("c1")
                #print((i,j),(i,j+1))

            try:
                if square[i+1][j+1]==1  and (i+1)>0 and (j+1)>0 and (i<n) and (j<m):
                    try:
                        neighbour_dictionary[(i,j)].setdefault([i,j],[]).append((i+1,j+1))
                    except KeyError:
                        neighbour_dictionary[(i, j)] = [(i+1, j+1)]
            except IndexError:
                pass
            #print("c2")
                #print((i,j),(i+1,j+1))

            try:

                if square[i+1][j-1]==1 and (i+1)>0 and (j-1)>0 and (i<n) and (j<m):

                    try:
                        neighbour_dictionary[(i, j)].setdefault([i,j],[]).append((i+1, j-1))
                    except KeyError:
                        neighbour_dictionary[(i, j)] = [(i+1,j-1)]
            except IndexError:
                pass
                #print("c3")
                #print((i,j),(i+1,j-1))

            try:

                if  square[i][j-1]==1  and (j-1)>0 and (i<n) and (j<m):

                    try:
                        neighbour_dictionary[(i, j)].setdefault([i,j],[]).append((i, j-1))
                    except KeyError:
                        neighbour_dictionary[(i, j)] = [(i, j + 1)]
            except IndexError:
                pass
                #print("c4")
                #print((i,j),(i,j-1))
            try:

                if square[i-1][j]==1 and (i-1)>0 and (j)>0 and (i<n) and (j<m):

                    try:
                        neighbour_dictionary[(i, j)].setdefault([i,j],[]).append((i-1, j))
                    except KeyError:
                        neighbour_dictionary[(i, j)] = [(i-1, j)]

            except IndexError:
                pass
                #print('c5')
            try:
                if square[i+1][j]==1  and (i+1)>0 and (j)>0 and (i<n) and (j<m):

                    try:
                        neighbour_dictionary[(i, j)].setdefault([i,j],[]).append((i+1, j))

                    except KeyError:
                        neighbour_dictionary[(i, j)] = [(i+1, j)]

            except IndexError:
                pass
                #print("c6")
                #print((i,j),(i+1,j))
            try:

                if square[i-1][j-1]==1  and (i-1)>0 and (j-1)>0:
                    try:
                        neighbour_dictionary[(i, j)].setdefault([i,j],[]).append((i-1, j-1))
                    except KeyError:
                        neighbour_dictionary[(i, j)] = [(i-1, j-1)]
            except IndexError:
                pass
                #print("c7")
                #print((i,j),(i-1,j-1))

print(neighbour_dictionary)

让我知道我在哪里做错了。

Answer 1

我已经解决了我自己的问题。边缘情况存在一个问题，这就是为什么它不起作用。我使用它们的条件是错误的。

谢谢大家的帮助。

如何在矩阵中查找特定类型的最近邻居

1 个答案: