Primary School Students Blue Bridge Cup Python Breakthrough | Polluted Seas

Learn Python from a baby! Record the questions in the Blue Bridge Cup Python learning and test preparation process, and record every moment.

Attached is a summary post: Pupils Blue Bridge Cup Python Breakthrough | Summary_COCOgsta's Blog-CSDN Blog

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【Description】

There is a sea area divided into N * M squares, some of which are polluted (indicated by 0), and some sea areas are not polluted (indicated by 1).

May I ask how many unpolluted independent sea areas are in this N * M sea area (unpolluted independent sea area means that the sea area is surrounded by polluted sea areas up, down, left, and right, and the sea areas other than N * M are all polluted) polluted seas)

For example: N=4, M=5, in the sea area of ​​4 * 5, the polluted sea area and the unpolluted sea area are as follows:

In this 4 * 5 sea area, 3 sea areas (green) are not polluted, because each block is surrounded by polluted sea areas.

【Enter description】

Input two positive integers N and M in the first line, N represents the row of the matrix grid, M represents the column of the matrix grid, and N and M are separated by a comma;

The second line starts to input N lines, each line has M numbers (numbers can only be 1 or 0, 1 means unpolluted sea area, 0 means polluted sea area)

【Output description】

Output an integer, indicating how many of the N * M sea areas are independent sea areas that are not polluted

【Sample input】

4,5

1,1,0,0,0

1,0,1,0,0

1,0,0,0,0

1,1,0,1,1

【Sample output】

3

【Code Explanation】

N, M = [int(i) for i in input().split(",")]
ls = [[int(i) for i in input().split(',')] for i in range(N)]

v = [[0] * M for i in range(N)] #记录探索结果

def dfs(x, y):
    v[x][y] = 1
    for  dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
        xx = x + dx
        yy = y + dy
        if 0<=xx<N and 0<=yy<M and ls[xx][yy] == 1 and v[xx][yy] == 0:
            dfs(xx, yy)

cnt = 0
for i in range(N):
    for j in range(M):
        if ls[i][j] == 1 and v[i][j] == 0:
            dfs(i, j)
            cnt += 1

print(cnt)
复制代码

【operation result】

4,5
1,1,0,0,0
1,0,1,0,0
1,0,0,0,0
1,1,0,1,1
3