代码随想录算法训练营第52天| 101. 孤岛的总面积、102. 沉没孤岛、103. 水流问题、104.建造最大岛屿
101. 孤岛的总面积
卡码题目链接：101. 孤岛的总面积
学习链接：代码随想录
题解：
法一：
count = 0
def dfs(grid,x,y):
global count
grid[x][y] = 0
count += 1
directions = [[1,0],[0,1],[-1,0],[0,-1]]
for i,j in directions:
next_x = x+i
next_y = y+j
if next_x < 0 or next_x >= len(grid) or next_y < 0 or next_y >= len(grid[0]):
continue
if grid[next_x][next_y] == 1:
dfs(grid,next_x,next_y)
def main():
# 因为这边不把count变成全局变量的话 count始终是初始化的0 不会有改变
global count
n,m = map(int,input().split())
grid = []
for _ in range(n):
grid.append(list(map(int,input().split())))
for i in range(n):
if grid[i][0] == 1:
dfs(grid,i,0)
if grid[i][m - 1] == 1:
dfs(grid, i, m - 1)
for j in range(m):
if grid[0][j] == 1:
dfs(grid,0,j)
if grid[n - 1][j] == 1:
dfs(grid, n - 1, j)
count = 0
for i in range(n):
for j in range(m):
if grid[i][j] == 1:
dfs(grid,i,j)
print(count)
if __name__ == "__main__":
main()Global变量使用：
x = 10 # 全局变量
def modify():
global x # 声明 x 是全局变量
x = 20 # 修改全局变量 x
print("Inside function:", x)
modify()
print("Outside function:", x)x可被修改
x = 10 # 全局变量
def modify():
x = 20 # 这里创建了一个局部变量 x
print("Inside function:", x)
modify()
print("Outside function:", x)x不可被修改
102. 沉没孤岛
卡码题目链接：102. 沉没孤岛
学习链接：代码随想录
题解：
法一：
def dfs(grid,x,y):
grid[x][y] = 2
directions = [[1,0],[0,1],[-1,0],[0,-1]]
for i,j in directions:
next_x = x+i
next_y = y+j
if next_x < 0 or next_x >= len(grid) or next_y < 0 or next_y >= len(grid[0]):
continue
if grid[next_x][next_y] == 2 or grid[next_x][next_y] == 0:
continue
dfs(grid,next_x,next_y)
def main():
n,m = map(int,input().split())
grid = []
for _ in range(n):
grid.append(list(map(int,input().split())))
for i in range(n):
if grid[i][0] == 1:
dfs(grid,i,0)
if grid[i][m-1] == 1:
dfs(grid,i,m-1)
for j in range(m):
if grid[0][j] == 1:
dfs(grid,0,j)
if grid[n-1][j] == 1:
dfs(grid,n-1,j)
for i in range(n):
for j in range(m):
if grid[i][j] == 2:
grid[i][j] = 1
elif grid[i][j] == 1:
grid[i][j] = 0
for row in grid:
print(" ".join(map(str,row)))
if __name__ == "__main__":
main()思路有点难 先把边上不是孤岛的转为2 然后再检查是2的变为1 是1的变为0
103. 水流问题
卡码题目链接：103. 水流问题
学习链接：代码随想录
题解：
法一：
first = set()
second = set()
def dfs(x,y,grid,visited,group):
if visited[x][y]:
return
directions = [[1,0],[0,1],[-1,0],[0,-1]]
visited[x][y] = True
group.add((x,y))
for i,j in directions:
next_x = x+i
next_y = y+j
if (0 <= next_x < len(grid)) and (0 <= next_y < len(grid[0])) and grid[next_x][next_y] >= grid[x][y]:
dfs(next_x,next_y,grid,visited,group)
def main():
global first
global second
n,m = map(int,input().split())
grid = []
for _ in range(n):
grid.append(list(map(int,input().split())))
visited = [[False] * m for _ in range(n)]
for i in range(n):
dfs(i,0,grid,visited,first)
for j in range(m):
dfs(0,j,grid,visited,first)
visited = [[False] * m for _ in range(n)]
for i in range(n):
dfs(i,m-1,grid,visited,second)
for j in range(m):
dfs(n-1,j,grid,visited,second)
# 取并集的写法
res = first.intersection(second)
for i,j in res:
print(f"{i} {j}")
if __name__ == "__main__":
main()
一个反转是把从高到低找边界 变成边界从低找高处
然后把边界分为左上和右下两部分 各自找到之后取相交。
104.建造最大岛屿
卡码题目链接：104. 建造最大岛屿
乐扣题目链接：827. 最大人工岛 - 力扣（LeetCode）
学习链接：代码随想录
题解：
卡码：
from collections import defaultdict
directions = [[1,0],[0,1],[-1,0],[0,-1]]
def dfs(grid,visited,x,y,num):
if visited[x][y]:
return
area = 1
visited[x][y] = True
grid[x][y] = num
for i,j in directions:
next_x = x+i
next_y = y+j
if (0 <= next_x < len(grid)) and (0 <= next_y < len(grid[0])) and grid[next_x][next_y] == "1":
area += dfs(grid,visited,next_x,next_y,num)
return area
def main():
n,m = map(int, input().strip().split())
grid = []
for _ in range(n):
grid.append(input().strip().split())
visited = [[False] * m for _ in range(n)]
rec = defaultdict(int)
# 编号
cnt = 2
for i in range(n):
for j in range(m):
if grid[i][j] == "1":
area = 0
rec[cnt] = dfs(grid,visited,i,j,cnt)
cnt += 1
res = 0
for i in range(n):
for j in range(m):
if grid[i][j] == "0":
max_island = 1
v = set()
for dx,dy in directions:
next_x = i+dx
next_y = j+dy
if (0 <= next_x < len(grid)) and (0 <= next_y < len(grid[0])) and grid[next_x][next_y] != "0" and grid[next_x][next_y] not in v:
max_island += rec[grid[next_x][next_y]]
v.add(grid[next_x][next_y])
res = max(res,max_island)
if res == 0:
res = max(rec.values()) if rec else 0
print(res)
if __name__ == "__main__":
main()
注意area的取值问题 是否能取到
这题太难了 好复杂 要操作的东西很多。
乐扣题解：
import collections
directions = [[-1, 0], [0, 1], [0, -1], [1, 0]]
class Solution:
def dfs(self,grid,visited,x,y,num):
if visited[x][y]:
return 0
visited[x][y] = True
area = 1
grid[x][y] = num
for i,j in directions:
next_x = i + x
next_y = j + y
if 0 <= next_x < len(grid) and 0 <= next_y < len(grid[0]) and grid[next_x][next_y] == 1:
area += self.dfs(grid,visited,next_x,next_y,num)
return area
def largestIsland(self, grid: List[List[int]]) -> int:
visited = [[False] * len(grid[0]) for _ in range(len(grid))]
rec = collections.defaultdict(int)
cnt = 2
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
area = 0
rec[cnt] = self.dfs(grid,visited,i,j,cnt)
cnt += 1
res = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 0:
max_island = 1
v = set()
for dx,dy in directions:
next_x = dx+i
next_y = dy+j
if 0 <= next_x < len(grid) and 0 <= next_y < len(grid[0]) and grid[next_x][next_y] != 0 and grid[next_x][next_y] not in v:
v.add(grid[next_x][next_y])
max_island += rec[grid[next_x][next_y]]
res = max(res,max_island)
if res == 0:
res = max(rec.values()) if rec else 0
return res