【代码随想录训练营】【Day 66】【图论-3】| 卡码 101-104
需强化知识点
- 103，104 优化思路
题目
101. 孤岛的总面积
- 此处 area 多余
python
def dfs(grid, x, y, area):
dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]]
m, n = len(grid), len(grid[0])
area[0] += 1
grid[x][y] = 0
for add_x, add_y in dirs:
next_x, next_y = x + add_x, y + add_y
if next_x < 0 or next_x >= m or next_y < 0 or next_y >= n:
continue
if grid[next_x][next_y]:
dfs(grid, next_x, next_y, area)
tmp = list(map(int, input().split()))
m, n = tmp[0], tmp[1]
grid = [[0] * n for _ in range(m)]
for i in range(m):
tmp = list(map(int, input().split()))
for j in range(n):
grid[i][j] = tmp[j]
for i in range(m):
if grid[i][0]:
dfs(grid, i, 0, [0])
if grid[i][n-1]:
dfs(grid, i, n-1, [0])
for j in range(n):
if grid[0][j]:
dfs(grid, 0, j, [0])
if grid[m-1][j]:
dfs(grid, m-1, j, [0])
cur = 0
for i in range(m):
for j in range(n):
if grid[i][j]:
cur += 1
print(cur) 102. 沉没孤岛
- 思路：从左右上下边界出发遍历，然后visited数组标记，最后 grid 为 1 且没被访问过的，即为孤岛
python
import collections
def bfs(grid, visited, x, y):
dirs = [[1, 0], [0, 1], [-1, 0], [0, -1]]
m, n = len(grid), len(grid[0])
que = collections.deque()
que.append([x, y])
visited[x][y] = True
while que:
tmp = que.popleft()
cur_x, cur_y = tmp[0], tmp[1]
for add_x, add_y in dirs:
next_x, next_y = cur_x + add_x, cur_y + add_y
if next_x < 0 or next_x >= m or next_y < 0 or next_y >= n:
continue
if grid[next_x][next_y] and not visited[next_x][next_y]:
que.append([next_x, next_y])
visited[next_x][next_y] = True
tmp = list(map(int, input().split()))
m, n = tmp[0], tmp[1]
grid = [[0] * n for _ in range(m)]
visited = [[False] * n for _ in range(m)]
for i in range(m):
tmp = list(map(int, input().split()))
for j in range(n):
grid[i][j] = tmp[j]
for i in range(m):
if grid[i][0]:
bfs(grid, visited ,i, 0)
if grid[i][n-1]:
bfs(grid, visited, i, n-1)
for j in range(n):
if grid[0][j]:
bfs(grid, visited, 0, j)
if grid[m-1][j]:
bfs(grid, visited, m-1, j)
for i in range(m):
for j in range(n):
if grid[i][j] and not visited[i][j]:
grid[i][j] = 0
for i in range(m):
for j in range(n):
print(grid[i][j], end=" ")
103. 水流问题
- 暴力法：直接每个位置 dfs，然后根据其最终是否能到达边界位置，返回布尔值
- 优化思路：从边界出发，逆流而上，最终不能被访问到的地方为结果
python
# def dfs(grid, visited, x, y):
# dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]]
# m, n = len(grid), len(grid[0])
# visited[x][y] = True
# for add_x, add_y in dirs:
# next_x, next_y = x + add_x, y + add_y
# if next_x < 0 or next_x >= m or next_y < 0 or next_y >= n:
# continue
# if grid[x][y] < grid[next_x][next_y]:
# continue
# if not visited[next_x][next_y]:
# dfs(grid, visited, next_x, next_y)
# def isResult(grid, x, y):
# m, n = len(grid), len(grid[0])
# visited = [[False] * n for _ in range(m)]
# dfs(grid, visited, x, y)
# first_result, second_result = False, False
# for i in range(m):
# if visited[i][0]:
# first_result = True
# if visited[i][n-1]:
# second_result = True
# for j in range(n):
# if visited[0][j]:
# first_result = True
# if visited[m-1][j]:
# second_result = True
# return first_result and second_result
# tmp = list(map(int, input().split()))
# m, n = tmp[0], tmp[1]
# grid = [[0] * n for _ in range(m)]
# for i in range(m):
# tmp = list(map(int, input().split()))
# for j in range(n):
# grid[i][j] = tmp[j]
# for i in range(m):
# for j in range(n):
# if isResult(grid, i, j):
# print("{} {}".format(i, j))
def dfs(grid, visited, x, y):
dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]]
m, n = len(grid), len(grid[0])
visited[x][y] = True
for add_x, add_y in dirs:
next_x, next_y = x + add_x, y + add_y
if next_x < 0 or next_x >= m or next_y < 0 or next_y >= n:
continue
# 等于不行
if grid[x][y] > grid[next_x][next_y]:
continue
if not visited[next_x][next_y]:
dfs(grid, visited, next_x, next_y)
tmp = list(map(int, input().split()))
m, n = tmp[0], tmp[1]
grid = [[0] * n for _ in range(m)]
for i in range(m):
tmp = list(map(int, input().split()))
for j in range(n):
grid[i][j] = tmp[j]
visited_first = [[False]*n for _ in range(m)]
visited_second = [[False]*n for _ in range(m)]
for i in range(m):
dfs(grid, visited_first, i, 0)
dfs(grid, visited_second, i, n-1)
for j in range(n):
dfs(grid, visited_first, 0, j)
dfs(grid, visited_second, m-1, j)
for i in range(m):
for j in range(n):
if visited_first[i][j] and visited_second[i][j]:
print("{} {}".format(i, j))
104. 建造最大岛屿
- 暴力法：直接每个为0的位置，dfs，记录其面积
- 优化思路：先记录每个岛屿的面积，并编号，然后 每个为0的位置，假设其为1，然后加上周围能访问到岛屿面积
- 注意周围访问岛屿的去重问题，以及为grid 0的情况
python
def dfs(grid, mask, x, y, count):
dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]]
m, n = len(grid), len(grid[0])
grid[x][y] = mask
count[0] += 1
for add_x, add_y in dirs:
next_x, next_y = x + add_x, y + add_y
if next_x < 0 or next_x >= m or next_y < 0 or next_y >= n:
continue
if grid[next_x][next_y] != 1 :
continue
dfs(grid, mask, next_x, next_y, count)
def main():
tmp = list(map(int, input().split()))
m, n = tmp[0], tmp[1]
grid = [[0] * n for _ in range(m)]
for i in range(m):
tmp = list(map(int, input().split()))
for j in range(n):
grid[i][j] = tmp[j]
mask = 2
isAllgrid = True
gridNum = {}
for i in range(m):
for j in range(n):
if grid[i][j] == 0:
isAllgrid = False
if grid[i][j] == 1:
count = [0]
dfs(grid, mask, i, j, count)
gridNum[mask] = count[0]
mask += 1
if isAllgrid:
print(m*n)
return
result = 0
dirs = [[0, 1], [0, -1], [1, 0], [-1, 0]]
for i in range(m):
for j in range(n):
if grid[i][j] == 0:
tmp = 1
visitedGrid = []
for add_x, add_y in dirs:
next_x, next_y = i + add_x, j + add_y
if next_x < 0 or next_x >= m or next_y < 0 or next_y >= n:
continue
if grid[next_x][next_y] not in visitedGrid and grid[next_x][next_y] != 0:
tmp += gridNum[grid[next_x][next_y]]
visitedGrid.append(grid[next_x][next_y])
result = max(result, tmp)
print(result)
main()