20240820 代码随想录 ｜ 图论 岛屿
98. 所有可达路径
深度优先搜索（dfs）和广度优先搜索（bfs）区别：
- dfs是可一个方向去搜，不到黄河不回头，直到遇到绝境了，搜不下去了，再换方向（换方向的过程就涉及到了回溯）。
- bfs是先把本节点所连接的所有节点遍历一遍，走到下一个节点的时候，再把连接节点的所有节点遍历一遍，搜索方向更像是广度，四面八方的搜索过程。
n, m = map(int, input().split())
print(' '.join(map(str, path)))
def dfs(graph, point, n, path, result):
if point == n:
result.append(path[:])
return
for i in range(1, n + 1):
if graph[point][i] == 1:
path.append(i)
dfs(graph, i, n, path, result)
path.pop()
def main():
n, m = map(int, input().split())
graph = [[0]*(n+1) for _ in range(n+1)]
for _ in range(m):
s, t = map(int, input().split())
graph[s][t] = 1
path = [1]
results = []
dfs(graph, 1, n, path, results)
if not results:
print(-1)
else:
for path in results:
print(' '.join(map(str, path)))
if __name__ == "__main__":
main()
99. 岛屿数量
dfs深度优先搜索：
direction = [[0,1],[0,-1],[-1,0],[1,0]]
def dfs(graph, i, j):
for d in direction:
x = i + d[0]
y = j + d[1]
if x >= 0 and x < len(graph) and y >= 0 and y < len(graph[0]) and graph[x][y] == 1:
graph[x][y] = 0
dfs(graph, x, y)
n,m = map(int, input().split())
graph = []
for _ in range(n):
graph.append(list(map(int, input().split())))
result = 0
for i in range(n):
for j in range(m):
if graph[i][j] == 1:
result += 1
dfs(graph, i, j)
graph[i][j] = 0
print(result)bfs广度优先搜索：
from collections import deque
direction = [[0,1],[0,-1],[-1,0],[1,0]]
def bfs(graph, i, j, visited):
que = deque()
que.append([i,j])
visited[i][j] = True
while que:
cur_x, cur_y = que.popleft()
for d in direction:
next_x = cur_x + d[0]
next_y = cur_y + d[1]
if next_x >= 0 and next_x < len(graph) and next_y >= 0 and next_y < len(graph[0]) and not visited[next_x][next_y] and graph[next_x][next_y] == 1:
que.append([next_x,next_y])
visited[next_x][next_y] = True
n,m = map(int, input().split())
graph = []
for _ in range(n):
graph.append(list(map(int, input().split())))
visited = [[False]*m for _ in range(n)]
result = 0
for i in range(n):
for j in range(m):
if graph[i][j] == 1 and not visited[i][j]:
result += 1
# visited[i][j] = True
bfs(graph, i, j, visited)
print(result)100. 岛屿的最大面积
bfs
这里注意count不能当成bfs的参数传，因为这样的话只是局部改变bfs内的count，主函数中count不受影响。
from collections import deque
directions = [[0,1],[0,-1],[1,0],[-1,0]]
def bfs(graph, x, y, visited):
que = deque()
que.append([x,y])
visited[x][y] = True
count = 1
while que:
cur_x, cur_y = que.popleft()
for d in directions:
next_x = cur_x + d[0]
next_y = cur_y + d[1]
if next_x >= 0 and next_x < len(graph) and next_y >= 0 and next_y < len(graph[0]) and not visited[next_x][next_y] and graph[next_x][next_y] == 1:
count += 1
visited[next_x][next_y] = True
que.append([next_x,next_y])
return count
n, m = map(int, input().split())
graph = []
for _ in range(n):
graph.append(list(map(int, input().split())))
visited = [[False]*m for _ in range(n)]
result = 0
for i in range(n):
for j in range(m):
if graph[i][j] == 1 and not visited[i][j]:
count = bfs(graph, i, j, visited)
result = max(result, count)
print(result)dfs：
注意这里，我没有用visited标记遍历过的点，而是直接将遍历过的1变成0，以避免dfs中的递归进入死循环。
directions = [[0,1],[0,-1],[1,0],[-1,0]]
count = 0
def dfs(graph, x, y):
global count
for d in directions:
next_x = x + d[0]
next_y = y + d[1]
if next_x >= 0 and next_x < len(graph) and next_y >= 0 and next_y < len(graph[0]) and graph[next_x][next_y] == 1:
count += 1
graph[next_x][next_y] = 0
dfs(graph, next_x, next_y)
n, m = map(int, input().split())
graph = []
for _ in range(n):
graph.append(list(map(int, input().split())))
result = 0
for i in range(n):
for j in range(m):
if graph[i][j] == 1:
count = 1
graph[i][j] = 0
dfs(graph, i, j)
result = max(result, count)
print(result)101. 孤岛的总面积
深搜dfs：
# count = 0
directions = [[0,1],[0,-1],[1,0],[-1,0]]
def dfs(graph, x, y):
# global count
# count += 1
graph[x][y] = 0
for d in directions:
next_x = x + d[0]
next_y = y + d[1]
if next_x >= 0 and next_x < len(graph) and next_y>=0 and next_ycount = 0
directions = [[0,1],[0,-1],[1,0],[-1,0]]
def dfs(graph, x, y):
global count
# 不能在这里加一，因为后面递归之前加了一次，递归的时候进来又加一次就重复了
# count += 1
graph[x][y] = 0
for d in directions:
next_x = x + d[0]
next_y = y + d[1]
if next_x >= 0 and next_x < len(graph) and next_y>=0 and next_y广搜bfs：
from collections import deque
count = 0
directions = [[0,1],[0,-1],[1,0],[-1,0]]
def bfs(graph, x, y):
que = deque()
que.append([x,y])
graph[x][y] = 0
global count
# 不能在这里加一，因为后面递归之前加了一次，递归的时候进来又加一次就重复了
# count += 1
while que:
x, y = que.popleft()
for d in directions:
next_x = x + d[0]
next_y = y + d[1]
if next_x >= 0 and next_x < len(graph) and next_y>=0 and next_y102. 沉没孤岛
dfs深度优先搜索
directions = [[0,1], [0,-1], [1,0], [-1,0]]
n,m = map(int, input().split())
g = []
for _ in range(n):
g.append(list(map(int, input().split())))
def dfs(g, x, y):
g[x][y] = 2
for d in directions:
next_x = x + d[0]
next_y = y + d[1]
if next_x>=0 and next_x=0 and next_y 0:
g[i][j] -= 1
for i in range(n):
print(" ".join(map(str, g[i]))) bfs广度优先搜索
from collections import deque
directions = [[0,1], [0,-1], [1,0], [-1,0]]
n,m = map(int, input().split())
g = []
for _ in range(n):
g.append(list(map(int, input().split())))
def dfs(g, x, y):
que = deque()
que.append([x,y])
g[x][y] = 2
while que:
x, y = que.popleft()
for d in directions:
next_x = x + d[0]
next_y = y + d[1]
if next_x>=0 and next_x=0 and next_y 0:
g[i][j] -= 1
for i in range(n):
print(" ".join(map(str, g[i])))