110. 字符串接龙

卡码网 110. 字符串接龙
代码随想录

广度优先第一个遇到的就是最短的。

from collections import deque
n = int(input().strip())start, end = input().split()strList = []
for _ in range(n):strList.append(input().strip())visited = [False] * n# 判断是否只有一个字符不同
def judge(str1, str2):count = 0for i in range(len(str1)):if str1[i] != str2[i]:count += 1return count == 1if start == end:print(0)exit()queue = deque()
queue.append([start, 1])while queue:str1, step = queue.popleft()if judge(str1, end):print(step+1)exit()for i in range(n):if visited[i] == False and judge(str1, strList[i]):visited[i] = Truequeue.append([strList[i], step+1])
print(0)

105.有向图的完全可达性

卡码网 105.有向图的完全可达性
代码随想录

从1出发，广度优先遍历，没遍历到的就是不可达的。

import collectionsn, k = map(int, input().split())visited = [False] * (n+1)
graph = collections.defaultdict(list)
# graph = collections.defaultdict(list)for _ in range(k):i,j = map(int, input().split())graph[i].append(j)def bfs(graph, visited, start):queue = []queue.append(start)while queue:node = queue.pop()visited[node] = Truefor i in range(len(graph[node])):if not visited[graph[node][i]]:queue.append(graph[node][i])
# 遍历                
bfs(graph, visited, 1)# 检查是否所有的节点都可达
for i in range(1, n+1):if visited[i] == False:print(-1)exit()
print(1)

106. 岛屿的周长

卡码网 106. 岛屿的周长
代码随想录

加一个水边框，一块陆地遇到有水的方向周长就可以加1了。

n, m = map(int, input().split())
graph = []for _ in range(n):graph.append(list(map(int, input().split())))visited = [[False]* (m+2) for _ in range(n+2)]# 给graph加上一个边框
graph.insert(0,[0]*m)
graph.append([0]*m)for i in range(n+2):graph[i].insert(0,0)graph[i].append(0)directions = [[-1,0], [1,0], [0,1], [0,-1]]def bfs(graph, visited, x, y):radius = 0queue = []visited[x][y] = Truequeue.append([x,y])while queue:cur_x, cur_y = queue.pop()for i, j in directions:new_x, new_y = cur_x+i, cur_y+j# 遍历if graph[new_x][new_y] == 1 and not visited[new_x][new_y]:queue.append([new_x, new_y])visited[new_x][new_y] = Trueif graph[new_x][new_y] == 0:radius += 1visited[new_x][new_y] = Truereturn radiusres = 0
for i in range(1, n+1):for j in range(1, m+1):if graph[i][j] == 1 and not visited[i][j]:res = bfs(graph, visited, i, j)
print(res)