【代码随想录算法训练营——Day52】图论——101.孤岛的总面积、102.沉没孤岛、103.水流问题、104.建造最大岛屿

卡码网题目链接
https://kamacoder.com/problempage.php?pid=1173
https://kamacoder.com/problempage.php?pid=1174
https://kamacoder.com/problempage.php?pid=1175
https://kamacoder.com/problempage.php?pid=1176

题解
101.孤岛的总面积
怎么不计算有边缘面积的岛屿呢？

最后只要重新遍历一遍地图就可以，因为是求面积。
在dfs遍历中，遇到海洋要跳过，这是一个错误点。
bfs很多语法不太熟悉，要照着写。

102.沉没孤岛
能不能把上一题用到的graph留下，对照着原始的graph将其孤岛变为0，再输出，这样的话代码几乎和上题一样。看看题解有没有什么更好的方法。

果然有更好的方法，就是标记2。

103.水流问题
遍历每一个点，看是否能到达第一组边界和第二组边界，同时每次遍历它周边的点时要看是否是坡度相等或更低的地形。还要一个visited数组，每次遍历时重新初始化为全False，避免重复遍历同一个点。
看了下题解，直接用visited的true痕迹记录到达第一边界和第二边界与否。

104.建造最大岛屿
暴力法，对每个点变成1，然后从该点开始bfs遍历计算当前的岛屿面积是多少，总的取最大值就行，但会不会和上一题一样是超时？看题解知道的，记住这样的算法时间复杂度是n^4。
优化方法看着有些难，c++和python的实现都是。题解的代码看不懂，建议多看。

代码

#101.孤岛的总面积
#dfs法
point = [[1, 0], [0, 1], [-1, 0], [0, -1]]
def dfs(graph, x, y):graph[x][y] = 0for 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]):continueif graph[nextx][nexty] == 0: continue;dfs(graph, nextx, nexty)if __name__ == "__main__":n, m = map(int, input().split())graph = []for i in range(n):graph.append(list(map(int, input().split())))for i in range(n):if graph[i][0] == 1:dfs(graph, i, 0)if graph[i][m - 1] == 1:dfs(graph, i, m - 1)for j in range(m):if graph[0][j] == 1:dfs(graph, 0, j)if graph[n - 1][j] == 1:dfs(graph, n - 1, j)result = 0for i in range(n):for j in range(m):if graph[i][j] == 1:result += 1print(result)#bfs法
from collections import deque
point = [[1, 0], [0, 1], [-1, 0], [0, -1]]
def bfs(graph, x, y):queue = deque()queue.append(x)queue.append(y)while queue:i = queue.popleft()j = queue.popleft()graph[i][j] = 0for t in range(4):nextx = i + point[t][0]nexty = j + point[t][1]if nextx < 0 or nextx >= len(graph) or nexty < 0 or nexty >= len(graph[0]):continueif graph[nextx][nexty] == 0: continue;queue.append(nextx)queue.append(nexty)if __name__ == "__main__":n, m = map(int, input().split())graph = []for i in range(n):graph.append(list(map(int, input().split())))for i in range(n):if graph[i][0] == 1:bfs(graph, i, 0)if graph[i][m - 1] == 1:bfs(graph, i, m - 1)for j in range(m):if graph[0][j] == 1:bfs(graph, 0, j)if graph[n - 1][j] == 1:bfs(graph, n - 1, j)result = 0for i in range(n):for j in range(m):if graph[i][j] == 1:result += 1print(result)

#102.沉没孤岛
#dfs法
point = [[0, 1], [1, 0], [0, -1], [-1, 0]]
def dfs(graph, x, y):graph[x][y] = 2for 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]):continueif graph[nextx][nexty] == 1:dfs(graph, nextx, nexty)if __name__ == "__main__":n, m = map(int, input().split())graph = []for i in range(n):graph.append(list(map(int, input().split())))for i in range(n):if graph[i][0] == 1:dfs(graph, i, 0)if graph[i][m - 1] == 1:dfs(graph, i, m - 1)for j in range(m):if graph[0][j] == 1:dfs(graph, 0, j)if graph[n - 1][j] == 1:dfs(graph, n - 1, j)for i in range(n):for j in range(m):if graph[i][j] == 1:graph[i][j] = 0for i in range(n):for j in range(m):if graph[i][j] == 2:graph[i][j] = 1for i in range(n):for j in range(m - 1):print(f"{graph[i][j]} ", end = '')print(graph[i][m - 1])#bfs法
from collections import deque
point = [[0, 1], [1, 0], [0, -1], [-1, 0]]
def bfs(graph, x, y):queue = deque()queue.append(x)queue.append(y)while queue:i = queue.popleft()j = queue.popleft()graph[i][j] = 2for t in range(4):nextx = i + point[t][0]nexty = j + point[t][1]if nextx < 0 or nextx >= len(graph) or nexty < 0 or nexty >= len(graph[0]):continueif graph[nextx][nexty] == 1:queue.append(nextx)queue.append(nexty)if __name__ == "__main__":n, m = map(int, input().split())graph = []for i in range(n):graph.append(list(map(int, input().split())))for i in range(n):if graph[i][0] == 1:bfs(graph, i, 0)if graph[i][m - 1] == 1:bfs(graph, i, m - 1)for j in range(m):if graph[0][j] == 1:bfs(graph, 0, j)if graph[n - 1][j] == 1:bfs(graph, n - 1, j)for i in range(n):for j in range(m):if graph[i][j] == 1:graph[i][j] = 0for i in range(n):for j in range(m):if graph[i][j] == 2:graph[i][j] = 1for i in range(n):for j in range(m - 1):print(f"{graph[i][j]} ", end = '')print(graph[i][m - 1])

#103.水流问题
#我写的不对的代码
point = [[0, 1], [1, 0], [-1, 0], [0, -1]]
def dfs(graph, visited, x, y, local_contact):visited[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]):continueif graph[nextx][nexty] <= graph[x][y] and visited[nextx][nexty] == False:if nextx == 0 or nexty == 0:local_contact[x][y][0] = Trueif nextx == n - 1 or nexty == m - 1:local_contact[x][y][1] = Truevisited[nextx][nexty] = Truedfs(graph, visited, nextx, nexty, local_contact)if __name__ == "__main__":n, m = map(int, input().split())graph = []for i in range(n):graph.append(list(map(int, input().split())))result = []local_contact = [[[False] * 2 for _ in range(m)] for _ in range(n)]for i in range(n):for j in range(m):visited = [[False] * m for _ in range(n)]dfs(graph, visited, i, j, local_contact)for i in range(n):for j in range(m):if local_contact[i][j][0] == True and local_contact[i][j][1] == True:result.append([i, j])for i in range(len(result)):print(f"{result[i][0]} {result[i][1]}")
#题解的暴力法，跟我的思路一样，实现不同，具体去看题解的代码，因为是c++不好放上来，而且这个思路超时了，时间复杂度是O(M^2*N*2)
#代码略
#优化代码
point = [[0, 1], [1, 0], [-1, 0], [0, -1]]
def dfs(graph, visited, x, y):visited[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]):continueif graph[nextx][nexty] >= graph[x][y] and visited[nextx][nexty] == False:#visited[nextx][nexty] = Truedfs(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())))result = []firstBorder = [[False] * m for _ in range(n)]secondBorder = [[False] * m for _ in range(n)]for i in range(n):dfs(graph, firstBorder, i, 0)dfs(graph, secondBorder, i, m - 1)for j in range(m):dfs(graph, firstBorder, 0, j)dfs(graph, secondBorder, n - 1, j)for i in range(n):for j in range(m):if firstBorder[i][j] == True and secondBorder[i][j] == True:result.append([i, j])for i in range(len(result)):print(f"{result[i][0]} {result[i][1]}")

#104.建造最大岛屿