蓝桥杯 全球变暖

题目https://www.lanqiao.cn/problems/178/learning/?page=1&first_category_id=1&sort=students_count&name=%E5%85%A8%E7%90%83%E5%8F%98%E6%9A%96

用flag=True表名这片岛屿已经满足了“不被淹没”的条件。

def dfs(grid,i,j):
    if not 0 <= i < len(grid) or not 0 <= j < len(grid[0]) or grid[i][j] != '#':
        return
    if cnt == 4:
        flag = True

    if flag == False:
        if grid[i+1][j] == '#' and grid[i-1][j] == '#' and grid[i][j+1] == '#' and grid[i][j-1]:
            cnt += 1

n = int(input())
grid = []
ans = 0
vis = [[0]*n for i in range(n)]
for i in range(n):
    nei = []
    nei = list(map(str, input().split(" ")))
    grid.append(nei)


def dfs(grid, i, j):
    global flag
    global vis
    if not 0 <= i < len(grid) or not 0 <= j < len(grid[0]) or grid[i][j] != '#' or vis[i][j] == 1:
        return
    vis[i][j] = 1
    if grid[i + 1][j] == '#' and grid[i - 1][j] == '#' and grid[i][j + 1] == '#' and grid[i][j - 1] == '#':
        flag = True
    dfs(grid, i - 1, j)
    dfs(grid, i + 1, j)
    dfs(grid, i, j + 1)
    dfs(grid, i, j - 1)


for i in range(n):
    for j in range(n):
        if grid[i][j] == '#' and vis[i][j] == 0:
            flag = False
            dfs(grid, i, j)
            if flag == False:
                ans += 1
print(ans)

题解 

为什么要在深搜之前加限制条件呢，因为可能是会有数组越界条件

import os
import sys

# 请在此输入您的代码
sys.setrecursionlimit(60000)
n = int(input())
mp = [list(input()) for i in range(n)]#海域照片
vis = [[0]*n for i in range(n)]
ans = 0#被淹没的岛屿的数量
def dfs(x,y):
    dic = [(0,1),(0,-1),(1,0),(-1,0)]
    global flag
    global vis
    global mp
    vis[x][y]=1
    if mp[x][y+1]=='#' and mp[x][y-1]=='#' and mp[x+1][y]=='#' and mp[x-1][y]=='#':
        flag=1
    for i in range(4):
        nx = x+dic[i][0]
        ny = y+dic[i][1]
        if vis[nx][ny]==0 and mp[nx][ny]=='#':
            dfs(nx,ny)


for i in range(n):
    for j in range(n):
        if vis[i][j]==0 and mp[i][j]=='#':
            flag=0
            dfs(i,j)
            if flag==0:
                ans+=1
print(ans)

看了题解之后的答案

n = int(input())
grid = []
ans = 0
vis = [[0] * n for i in range(n)]
# for i in range(n):
#     nei = []
#     nei = list(map(str, input().split(" ")))
#     grid.append(nei)
grid = [list(input()) for i in range(n)]


def dfs(grid, i, j):
    dic = [[0, 1], [0, -1], [1, 0], [-1, 0]]
    global flag
    global vis
    if not 0 <= i < len(grid) or not 0 <= j < len(grid[0]) or grid[i][j] != '#' or vis[i][j] == 1:
        return
    vis[i][j] = 1
    if grid[i + 1][j] == '#' and grid[i - 1][j] == '#' and grid[i][j + 1] == '#' and grid[i][j - 1] == '#':
        flag = True
    for k in range(4):
        nx = i + dic[k][0]
        ny = j + dic[k][1]
        if vis[nx][ny] == 0 and grid[nx][ny] == '#':
            dfs(grid, nx, ny)

for i in range(n):
    for j in range(n):
        if grid[i][j] == '#' and vis[i][j] == 0:
            flag = False
            dfs(grid, i, j)
            if not flag:
                ans += 1
print(ans)

差别就是sys.setrecursionlimit(60000)

另一个比较好看的题解

# 连通性判断:
# 连通性问题,计算步骤:
# 遍历一个连通块(找到这个连通块中所有的'#',标记已经搜过,不用再搜)
# 再遍历下一个连通块......
# 遍历完所有连通块,统计有多少个连通块。
import sys

sys.setrecursionlimit(1000000)

# 因为这个只是判断我当前这个x,y是不是高地，并不是DFS搜索结束的条件，如果加return，那就有可能导致有陆地没有被搜索到过 
# 这个DFS函数的目的就是去搜索所有陆地，没有用到回溯的功能
def dfs(x, y):
    d = [(0, -1), (0, 1), (-1, 0), (1, 0)]
    global flag
    vis[x][y] = 1
    if mp[x][y - 1] == '#' and mp[x][y + 1] == '#' and mp[x - 1][y] == '#' and mp[x + 1][y] == '#':
        flag = 1
    for i in range(4):
        nx = x + d[i][0]
        ny = y + d[i][1]
        if vis[nx][ny] == 0 and mp[nx][ny] == '#':
            dfs(nx, ny)
n = int(input())
mp = []
for i in range(n):
    mp.append(list(input()))  # mp二维数组存图
# mp = [[''] * n for i in range(n)]
# for i in range(n):
#     mp[i] = list(input())
vis = [[0] * n for i in range(n)]  # vis数组用于判断图上的点是否走过,1是走过,0是没有走过
ans = 0
for i in range(n):
    for j in range(n):                             # 遍历每一个点
        if vis[i][j] == 0 and mp[i][j] == '#':     # 如果这个点没有走过,并且这个点是岛屿
            flag = 0                               # flag 是判断这个岛屿是不是高地,flag = 0不是,flag = 1是
            dfs(i, j)
            if flag == 0:                          # 如果这个岛屿被吞没,ans += 1
                ans += 1
print(ans)

更加完整的

import sys

sys.setrecursionlimit(1000000)
# grid = [list(input()) for i in range(n)]
# 0<= xx < len(grid) and 0 <= yy < len(grid[0]) and

def dfs(x, y):
    d = [(0, -1), (0, 1), (-1, 0), (1, 0)]
    global flag
    vis[x][y] = 1
    if mp[x][y - 1] == '#' and mp[x][y + 1] == '#' and mp[x - 1][y] == '#' and mp[x + 1][y] == '#':
        flag = 1
    for i in range(4):
        nx = x + d[i][0]
        ny = y + d[i][1]
        if vis[nx][ny] == 0 and mp[nx][ny] == '#' and 0<= nx < len(mp) and 0 <= ny < len(mp[0]):
            dfs(nx, ny)
n = int(input())
mp = [list(input()) for i in range(n)]



vis = [[0] * n for i in range(n)]
ans = 0
for i in range(n):
    for j in range(n):
        if vis[i][j] == 0 and mp[i][j] == '#':
            flag = 0
            dfs(i, j)
            if flag == 0:
                ans += 1
print(ans)