问题描述

小蓝有一个 30 行 60 列的数字矩阵，矩阵中的每个数都是 0 或 1 。

110010000011111110101001001001101010111011011011101001111110010000000001010001101100000010010110001111100010101100011110 001011101000100011111111111010000010010101010111001000010100 101100001101011101101011011001000110111111010000000110110000 010101100100010000111000100111100110001110111101010011001011 010011011010011110111101111001001001010111110001101000100011 101001011000110100001101011000000110110110100100110111101011 101111000000101000111001100010110000100110001001000101011001 001110111010001011110000001111100001010101001110011010101110 001010101000110001011111001010111111100110000011011111101010 011111100011001110100101001011110011000101011000100111001011 011010001101011110011011111010111110010100101000110111010110 001110000111100100101110001011101010001100010111110111011011 111100001000001100010110101100111001001111100100110000001101 001110010000000111011110000011000010101000111000000110101101 100100011101011111001101001010011111110010111101000010000111 110010100110101100001101111101010011000110101100000110001010 110101101100001110000100010001001010100010110100100001000011 100100000100001101010101001101000101101000000101111110001010 101101011010101000111110110000110100000010011111111100110010 101111000100000100011000010001011111001010010001010110001010 001010001110101010000100010011101001010101101101010111100101 001111110000101100010111111100000100101010000001011101100001 101011110010000010010110000100001010011111100011011000110010 011110010100011101100101111101000001011100001011010001110011 000101000101000010010010110111000010101111001101100110011100 100011100110011111000110011001111100001110110111001001000111 111011000110001000110111011001011110010010010110101000011111 011110011110110110011011001011010000100100101010110000010011 010011110011100101010101111010001001001111101111101110011101

如果从一个标为 1 的位置可以通过上下左右走到另一个标为 1 的位置，则称两个位置连通。与某一个标为 1 的位置连通的所有位置（包括自己）组成一个连通分块。

请问矩阵中最大的连通分块有多大？

答案提交

这是一道结果填空的题，你只需要算出结果后提交即可。本题的结果为一个整数，在提交答案时只填写这个整数，填写多余的内容将无法得分。

def dfs(x,y):d=[(0,1),(0,-1),(1,0),(-1,0)]global ans,visans+=1vis[x][y]=1  # 搜过标记为 1for i in range(4):nx=x+d[i][0]ny=y+d[i][1]if nx<0 or nx>29 or ny<0 or ny>59: #超出范围结束continueif mp[nx][ny]==1 and vis[nx][ny]==0:vis[nx][ny]=1dfs(nx,ny)
mp=[]
ans=0
a=0
vis=[[0]*60 for _ in range(30)]  #用来标记
for _ in range(30):mp.append([int(x) for x in input()])
for i in range(30):for j in range(60):if vis[i][j]==0 and mp[i][j]==1:dfs(i,j)a=max(a,ans)ans=0
print(a)

注意： 当设置范围的时候，要设置 nt 和 ny 的范围，而不是 x 、y 的范围，否则会报错（列表索引超出范围）