代码随想录算法营Day57 ｜ 孤岛的总面积，沉没孤岛，水流问题，建造最大岛屿
孤岛的总面积
这道题先将靠近边界线上的岛屿都放置为0，然后再用深度或者广度搜索算法去计算剩余的孤岛总面积
count = 0
position = [[1, 0], [0, 1], [-1, 0], [0, -1]]
def dfs(matrix,x,y):
global count
matrix[x][y] = 0
count += 1
for i,j in position:
next_x = x + i
next_y = y + j
if next_x < 0 or next_y < 0 or next_x >= len(matrix) or next_y >= len(matrix[0]):
continue
if matrix[next_x][next_y] == 1:
dfs(matrix,next_x,next_y)
N,M = map(int,input().split())
matrix = []
for _ in range(N):
matrix.append(list(map(int,input().split())))
for i in range(N):
if matrix[i][0] == 1:
dfs(matrix,i,0)
if matrix[i][M-1] == 1:
dfs(matrix,i,M-1)
for j in range(M):
if matrix[0][j] == 1:
dfs(matrix,0,j)
if matrix[N-1][j] == 1:
dfs(matrix,N-1,j)
count=0
for i in range(N):
for j in range(M):
if matrix[i][j] == 1:
dfs(matrix,i,j)
print(count)
沉没孤岛
这道题思路和上面一样，不过这里需要将边界上的岛屿变成2。那么此时矩阵当中就只有靠近边界上的2以及孤岛1。这时候只需要将所有数值为2或者为1的都减去1即可。
position = [[1, 0], [0, 1], [-1, 0], [0, -1]]
def dfs(matrix,x,y):
matrix[x][y] = 2
for i,j in position:
next_x = x + i
next_y = y + j
if next_x < 0 or next_y < 0 or next_x >= len(matrix) or next_y >= len(matrix[0]):
continue
if matrix[next_x][next_y] == 1:
dfs(matrix,next_x,next_y)
N,M = map(int,input().split())
matrix = []
for _ in range(N):
matrix.append(list(map(int,input().split())))
for i in range(N):
if matrix[i][0] == 1:
dfs(matrix,i,0)
if matrix[i][M-1] == 1:
dfs(matrix,i,M-1)
for j in range(M):
if matrix[0][j] == 1:
dfs(matrix,0,j)
if matrix[N-1][j] == 1:
dfs(matrix,N-1,j)
count=0
for i in range(N):
for j in range(M):
if matrix[i][j] == 1 or matrix[i][j] == 2:
matrix[i][j] -= 1
for l in matrix:
print(" ".join(map(str,l)))
水流问题
现有一个 N × M 的矩阵，每个单元格包含一个数值，这个数值代表该位置的相对高度。矩阵的左边界和上边界被认为是第一组边界，而矩阵的右边界和下边界被视为第二组边界。
矩阵模拟了一个地形，当雨水落在上面时，水会根据地形的倾斜向低处流动，但只能从较高或等高的地点流向较低或等高并且相邻（上下左右方向）的地点。我们的目标是确定那些单元格，从这些单元格出发的水可以达到第一组边界和第二组边界。
这道题我们从各边界逆着往内去找可达的路径，然后取这两个可达路径的交集即可。
N,M = map(int,input().split())
matrix = []
for _ in range(N):
matrix.append(list(map(int,input().split())))
first = set()
second = set()
directions = [[-1, 0], [0, 1], [1, 0], [0, -1]]
def dfs(i,j,grid,visited,side):
if visited[i][j]:
return
visited[i][j] = True
side.add((i,j))
for x,y in directions:
new_x = x + i
new_y = y + j
if 0<= new_x < N and 0<= new_y < M and grid[i][j] <= grid[new_x][new_y]:
dfs(new_x,new_y,grid,visited,side)
visited = [[False] * M for _ in range(N)]
for i in range(N):
dfs(i,0,matrix,visited,first)
for j in range(M):
dfs(0,j,matrix,visited,first)
visited = [[False] * M for _ in range(N)]
for i in range(N):
dfs(i,M-1,matrix,visited,second)
for j in range(M):
dfs(N-1,j,matrix,visited,second)
res = first & second
for x,y in res:
print(f"{x} {y}")
建造最大岛屿
这道题先将各个岛用不同的数字标示出来，并计算每个岛的面积，而后遍历每个海水记录可以由该海水块连接起来的陆地的最大面积。
N,M = map(int,input().split())
matrix = []
for _ in range(N):
matrix.append(list(map(int,input().split())))
directions = [[-1, 0], [0, 1], [0, -1], [1, 0]]
area = 0
def dfs(i,j,grid,visited,num):
global area
if visited[i][j]:
return
visited[i][j] = True
grid[i][j] = num
area += 1
for x, y in directions:
new_x = x + i
new_y = y + j
if 0 <= new_x < N and 0 <= new_y < M and grid[new_x][new_y] == 1:
dfs(new_x,new_y,grid,visited,num)
cnt = 2
visited = [[False] * M for _ in range(N)]
rec = {}
for i in range(N):
for j in range(M):
if matrix[i][j] == 1:
area = 0
dfs(i,j,matrix,visited,cnt)
rec[cnt] = area
cnt += 1
res = 0
for i in range(N):
for j in range(M):
if matrix[i][j] == 0:
max_island = 1
v = set()
for x, y in directions:
new_x = x + i
new_y = y + j
if 0 <= new_x < N and 0 <= new_y < M and matrix[new_x][new_y] != 0 and matrix[new_x][new_y] not in v:
max_island += rec[matrix[new_x][new_y]]
v.add(matrix[new_x][new_y])
res = max(res,max_island)
if res == 0:
print(max(rec.values()))
else:
print(res)