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【代码随想录算法训练营——Day53】图论——110.字符串接龙、105.有向图的完全可达性、106.岛屿的周长

news 2025/10/31 10:23:36

卡码网题目链接
https://kamacoder.com/problempage.php?pid=1183
https://kamacoder.com/problempage.php?pid=1177
https://kamacoder.com/problempage.php?pid=1178

题解
110.字符串接龙
因为接连做两天的题，有点放弃治疗。感觉这题不会。
看题解，原来要画图。关于做题原理有很多细节，看题解。

105.有向图的完全可达性
用邻接矩阵存储图，从1开始出发用一次dfs，同时用visited数组记录是否可达，最后判断visited是否都为true即可。
在这里插入图片描述

106.岛屿的周长
相当于是统计相邻的1的个数。我这份dfs代码怎么改都不对，不知道问题在哪里。
在这里插入图片描述
终于找到原因了，原来是双层循环没有跳出。

代码

#110.字符串接龙

在这里插入图片描述

#105.有向图的完全可达性
def dfs(graph, visited, y):visited[y] = Truefor i in range(1, len(graph)):if graph[y][i] == 1 and visited[i] == False:dfs(graph, visited, i)if __name__ == "__main__":n, m = map(int, input().split())graph = [[0] * (n + 1) for _ in range(n + 1)]for i in range(m):s, t = map(int, input().split())graph[s][t] = 1visited = [False] * (n + 1)dfs(graph, visited, 1)flag = Truefor i in range(1, n + 1):if visited[i] == False:flag = Falseif flag:print(1)else:print(-1)

#106.岛屿的周长
#错误代码
point = [[1, 0], [0, 1], [-1, 0], [0, -1]]
result = 0
def dfs(graph, visited, x, y):global resultvisited[x][y] = Truefor i in range(4):nextx = x + point[i][0]nexty = y + point[i][1]if nextx < 0 or nextx >= len(graph) or nexty < 0 or nexty >= len(graph[0]):result += 1elif graph[nextx][nexty] == 0:result += 1elif graph[nextx][nexty] == 1 and visited[nextx][nexty] == False:dfs(graph, visited, nextx, nexty)if __name__ == "__main__":n, m = map(int, input().split())graph = []for i in range(n):graph.append(list(map(int, input().split())))visited = [[False] * m for _ in range(n)]flag = Truefor i in range(n):for j in range(m):if graph[i][j] == 1:dfs(graph, visited, i, j)flag = Falsebreakif not flag:breakprint(result)#deeoseek加调试代码
point = [[1, 0], [0, 1], [-1, 0], [0, -1]]
result = 0def dfs(graph, visited, x, y):global resultvisited[x][y] = Trueprint(f"访问格子({x},{y})")for i in range(4):nextx = x + point[i][0]nexty = y + point[i][1]direction = ["下", "右", "上", "左"][i]if nextx < 0 or nextx >= len(graph) or nexty < 0 or nexty >= len(graph[0]):result += 1print(f"  {direction}方向: 地图边界，周长+1 → {result}")elif graph[nextx][nexty] == 0:result += 1print(f"  {direction}方向: 遇到水域，周长+1 → {result}")elif graph[nextx][nexty] == 1 and not visited[nextx][nexty]:print(f"  {direction}方向: 递归到({nextx},{nexty})")dfs(graph, visited, nextx, nexty)if __name__ == "__main__":n, m = 5, 5graph = [[0, 0, 0, 0, 0],[0, 1, 0, 1, 0], [0, 1, 1, 1, 0],[0, 1, 1, 1, 0],[0, 0, 0, 0, 0]]visited = [[False] * m for _ in range(n)]# 从(1,1)开始dfs(graph, visited, 1, 1)print(f"最终结果: {result}")