代码随想录算法训练营day51| 卡码网99.岛屿数量卡码网100.岛屿的最大面积

学习资料：https://www.programmercarl.com/kamacoder/0099.岛屿的数量深搜.html#思路

深度优先搜索和广度优先搜索
今天用的邻接矩阵

学习记录：
卡码网99.岛屿数量（深搜or广搜；用一个自己设计的二维矩阵来控制节点的移动方向：上下左右）

点击查看代码

from collections import deque
direction = [[1,0], [0,1], [-1,0], [0, -1]]  # 上，右，下，左

def bfs(visited, grid, x, y):
    """广搜"""
    que = deque([])
    que.append([x, y])
    while que:
        cur_x, cur_y = que.popleft()
        for i, j in direction:
            next_x = cur_x + i
            next_y = cur_y + j
            if next_x<0 or next_x>=len(grid) or next_y<0 or next_y>=len(grid[0]):
                continue
            if not visited[next_x][next_y] and grid[next_x][next_y] == 1:
                visited[next_x][next_y] = True
                que.append([next_x, next_y])
    


def dfs(visited, grid, x, y):
    """深搜"""
    for i, j in direction:
        next_x = x+i
        next_y = y + j
        # 下标越界，跳过 (这一小片区域的边界)
        if next_x<0 or next_x>=len(grid) or next_y<0 or next_y>=len(grid[0]):
            continue
        # 遇到为访问的陆地，标记并使用深搜
        if not visited[next_x][next_y] and grid[next_x][next_y]==1:
            visited[next_x][next_y] = True
            dfs(visited, grid, next_x, next_y)

if __name__ == "__main__":
    n, m = map(int, input().split())
    
    # 构造邻接矩阵
    grid = []
    for i in range(n):
        grid.append(list(map(int, input().split())))
    
    # 构造访问表，若以访问则为True
    visited = [[False]*m for _ in range(n)]
    
    # 岛屿数量
    res = 0
    
    for i in range(n):
        for j in range(m):
            # 如果当前是陆地，且未被访问过，说明找到了一片新的陆地，标记该访问情况，深搜找这片的范围
            if grid[i][j] == 1 and not visited[i][j]:
                res += 1
                visited[i][j] = True
                bfs(visited, grid, i, j)   # dfs(visited, grid, i, j) 也可以
    
    print(res)

卡码网100.岛屿的最大面积（深搜法；给前面这道题的基础上，遍历每片岛屿时，要记录每个陆地值得到岛屿面积）

点击查看代码

directions = [[1,0],[0,1],[-1,0],[0,-1]]
count = 0

def dfs(visited, grid, x, y):
    global count   # 设置全局变量
    for i,j in directions:
        cur_x = x + i
        cur_y = y + j
        if cur_x<0 or cur_x>=len(grid) or cur_y<0 or cur_y>=len(grid[0]):
            continue
        if not visited[cur_x][cur_y] and grid[cur_x][cur_y]==1:
            visited[cur_x][cur_y] = True
            count += 1
            dfs(visited, grid, cur_x, cur_y)



n, m = map(int, input().split())

grid = []
for i in range(n):
    grid.append(list(map(int, input().split())))
    
visited = [[False]*m for _ in range(n)]

result = 0     # 记录count的最大值

for i in range(n):
    for j in range(m):
        if grid[i][j]==1 and not visited[i][j]:
            count = 1
            visited[i][j] = True
            dfs(visited, grid, i, j)
            result = max(result, count)

print(result)