【代码随想录算法训练营第五十八天|卡码网101.孤岛的总面积、102.沉没孤岛、103.水流问题、104.建造最大岛屿】

文章目录

101.孤岛的总面积
102.沉没孤岛
103.水流问题
- 正向逻辑
- 反向逻辑
104.建造最大岛屿

101.孤岛的总面积

可以把最外围的都检查一遍是否有为1的，有的话就把他接壤的全变成海，然后正常算面积。也可以看岛屿是否有靠边的位置，有的话该岛面积不计算在总面积中。

direction = [[1, 0], [0, 1], [-1, 0], [0, -1]]
def bfs(cur, islands, visited):N = len(islands)M = len(islands[0])st = [cur]area = 1flag = Falseif cur[0]==0 or cur[0]==N-1 or cur[1]==0 or cur[1]==M-1:flag = Truewhile st:cur = st.pop()for d in direction:x = d[0] + cur[0]y = d[1] + cur[1]if x < 0 or x >= N or y < 0 or y >= M:continueif islands[x][y] == 1 and visited[x][y] == False:if x==0 or x==N-1 or y==0 or y==M-1:flag = Truearea += 1st.append([x, y])visited[x][y] = Trueif flag:return 0else:return area
if __name__ == '__main__':area = 0NM = input().split()N, M = int(NM[0]), int(NM[1])islands = [[0] * M for _ in range(N)]for i in range(N):lands = input().split()for j in range(M):islands[i][j] = int(lands[j])visited = [[False] * M for _ in range(N)]for i in range(N):for j in range(M):if (islands[i][j] == 1) and (visited[i][j]==False):visited[i][j] = Truetmp = bfs([i,j], islands, visited)# print([i, j], tmp)area += tmpprint(area)

102.沉没孤岛

这题就是从外圈找，找到未访问且为1的就把接壤的在新的岛屿图上标注为1。

direction = [[1, 0], [0, 1], [-1, 0], [0, -1]]def bfs(cur, islands, visited, new_islands):N = len(islands)M = len(islands[0])st = [cur]while st:cur = st.pop()for d in direction:x = d[0] + cur[0]y = d[1] + cur[1]if x < 0 or x >= N or y < 0 or y >= M:continueif islands[x][y] == 1 and visited[x][y] == False:st.append([x, y])visited[x][y] = Truenew_islands[x][y] = 1
if __name__ == '__main__':area = 0NM = input().split()N, M = int(NM[0]), int(NM[1])islands = [[0] * M for _ in range(N)]for i in range(N):lands = input().split()for j in range(M):islands[i][j] = int(lands[j])visited = [[False] * M for _ in range(N)]new_islands = [[0] * M for _ in range(N)]for j in range(M):if islands[0][j] == 1 and not visited[0][j]:new_islands[0][j] = 1bfs([0, j], islands, visited, new_islands)if islands[N-1][j] == 1 and not visited[N-1][j]:new_islands[N-1][j] = 1bfs([N-1, j], islands, visited, new_islands)for i in range(1, N-1):if islands[i][0] == 1 and not visited[i][0]:new_islands[i][0] = 1bfs([i, 0], islands, visited, new_islands)if islands[i][M-1] == 1 and not visited[i][M-1]:new_islands[i][M-1] = 1bfs([i, M-1], islands, visited, new_islands)for island in new_islands:print(' '.join(map(str, island)))

103.水流问题

正向逻辑

正向逻辑就是对每一块去看他能流到哪里，如果同时能流到第一边界和第二边界那就输出。但是会超时。

def dfs(cur, grid, visited):if visited[cur[0]][cur[1]]:returnvisited[cur[0]][cur[1]] = TrueN = len(grid)M = len(grid[0])directions = [[1,0], [-1,0],[0,1], [0,-1]]for d in directions:x = cur[0] + d[0]y = cur[1] + d[1]if x < 0 or x >= N or y < 0 or y >= M:continueif grid[x][y] <= grid[cur[0]][cur[1]]:dfs([x, y], grid, visited)def flow2Borad(cur, grid):N = len(grid)M = len(grid[0])visited = [[False] * M for _ in range(N)]dfs(cur, grid, visited)isFirst = FalseisSec = Falsefor i in range(M):if visited[0][i]:isFirst = Truebreakif not isFirst:for i in range(N):if visited[i][0]:isFirst = Truebreakfor i in range(M):if visited[N-1][i]:isSec = Truebreakif not isSec:for i in range(N):if visited[i][M-1]:isSec = Truebreak   if isSec and isFirst:return Trueelse:return False 
if __name__ == '__main__':area = 0NM = input().split()N, M = int(NM[0]), int(NM[1])islands = [[0] * M for _ in range(N)]for i in range(N):lands = input().split()for j in range(M):islands[i][j] = int(lands[j])for i in range(N):for j in range(M):if flow2Borad([i,j], islands):print(str(i), ' ',str(j))

反向逻辑

对第一边界和第二边界的格子往回推导，退出哪些格子能够流到第一边界和第二边界，如果一个格子能流到第一边界，并且可以流到第二边界那就输出他。

def dfs(cur, islands, visited):if visited[cur[0]][cur[1]]:returnN = len(islands)M = len(islands[0])visited[cur[0]][cur[1]] = Truedirections = [[1,0], [-1,0], [0,1], [0,-1]]for d in directions:x = cur[0] + d[0]y = cur[1] + d[1]if x < 0 or x >= N or y < 0 or y >= M:continueif islands[x][y] < islands[cur[0]][cur[1]]:continuedfs([x,y], islands, visited)if __name__ == '__main__':N, M = map(int, input().split())islands = [[0] * M for _ in range(N)]for i in range(N):lands = input().split()for j in range(M):islands[i][j] = int(lands[j])isFirst_Borad = [[False] * M for _ in range(N)]isSec_Board = [[False] * M for _ in range(N)]for i in range(N):dfs([i, 0], islands, isFirst_Borad)dfs([i, M-1], islands, isSec_Board)for j in range(M):dfs([0, j], islands, isFirst_Borad)dfs([N-1, j], islands, isSec_Board)for i in range(N):for j in range(M):if isSec_Board[i][j] and isFirst_Borad[i][j]:print(str(i)+' '+str(j))

104.建造最大岛屿

有点麻烦的这题，在前面的基础上还要再做一个列表来对岛屿进行标号。在计算完所有岛屿面积后做出一个岛屿标号和面积的字典，然后遍历所有的位置，如果是海洋，则在他四周找岛屿，没找到一个不同标号的岛屿就加到结果中，最后选最大的一个岛屿面积和。（但是题目说没有岛输出0，但是答案又是1，就是默认没有岛屿就翻转一块位置变成岛屿

direction = [[1, 0], [0, 1], [-1, 0], [0, -1]]def bfs(cur, islands, visited, mark, num):visited[cur[0]][cur[1]] = Truemark[cur[0]][cur[1]] = numst = [cur]area = 1while st:cur = st.pop()for d in direction:x = d[0] + cur[0]y = d[1] + cur[1]if x < 0 or x >= N or y < 0 or y >= M:continueif islands[x][y] == 1 and visited[x][y] == False:st.append([x, y])area += 1visited[x][y] = Truemark[x][y] = numreturn areaif __name__ == '__main__':N, M = map(int, input().split())islands = [[0] * M for _ in range(N)]for i in range(N):lands = input().split()for j in range(M):islands[i][j] = int(lands[j])visited = [[False] * M for _ in range(N)]mark = [[0] * M for _ in range(N)]num = 1islandArea = {}for i in range(N):for j in range(M):if (islands[i][j] == 1) and (visited[i][j]==False):area = bfs([i,j], islands, visited, mark, num)islandArea[num] = areanum += 1directions = [[1,0], [-1,0], [0,1], [0,-1]]if islandArea:result = max([value for value in islandArea.values()])for i in range(N):for j in range(M):if islands[i][j] == 0:temp = 1pre = []for d in directions:nextx = i + d[0]nexty = j + d[1]if nextx < 0 or nextx >= N or nexty < 0 or nexty >= M:continue                    if islands[nextx][nexty] == 1 and not mark[nextx][nexty] in pre:temp += islandArea[mark[nextx][nexty]]pre.append(mark[nextx][nexty])result = max(result, temp)print(result)else:print(1)