岛屿问题

 

 

 

 

python3:

lst = []
for q in range(4):
    lst.append(list(map(eval, list(input("")))))


mark = 2
print(lst[3][4])
dict0 = {}
#递归标记周边陆地
def markAround(x,y,mark):
    if (x+1)<=3 and lst[x+1][y] ==1:
        lst[x + 1][y] = mark
        dict0[mark] = dict0[mark]+2
        markAround(x+1,y,mark)
    if (x-1)>=0 and lst[x-1][y] ==1:
        lst[x - 1][y] = mark
        dict0[mark] = dict0[mark] + 2
        markAround(x-1,y,mark)
    if (y+1)<=4 and lst[x][y+1] ==1:
        lst[x][y+1] = mark
        dict0[mark] = dict0[mark] + 2
        markAround(x,y+1,mark)
    if (y-1)>=0 and lst[x][y-1] ==1:
        lst[x][y-1] = mark
        dict0[mark] = dict0[mark] + 2
        markAround(x,y-1,mark)

for x in range(4):
    for y in range(5):
        #若找到一块岛屿,就++count
        if lst[x][y]==1:
            lst[x][y] = mark
            dict0[mark] = 2
            markAround(x,y,mark)
            mark += 1

    print(lst)

# for i in range(4):
#     for j in range(5):
#         if lst[i][j]==0:
#             continue
#         if lst[i][j] in dict0.keys():
#             dict0[lst[i][j]] = dict0[lst[i][j]] + 2
#         if lst[i][j] not in dict0.keys():
#             dict0[lst[i][j]] = 2
print(dict0)
print(len(dict0))

 

 

TODO ： 优化效率