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 = 1
    flag = False
    if cur[0]==0 or cur[0]==N-1 or cur[1]==0 or cur[1]==M-1:
        flag = True
    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:
                continue
            if islands[x][y] == 1 and visited[x][y] == False:
                if x==0 or x==N-1 or y==0 or y==M-1:
                    flag = True
                area += 1
                st.append([x, y])
                visited[x][y] = True
    if flag:
        return 0
    else:
        return area
if __name__ == '__main__':
    area = 0
    NM = 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] = True
                tmp = bfs([i,j], islands, visited)
                # print([i, j], tmp)
                area += tmp
    print(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:
                continue
            if islands[x][y] == 1 and visited[x][y] == False:
                st.append([x, y])
                visited[x][y] = True
                new_islands[x][y] = 1
if __name__ == '__main__':
    area = 0
    NM = 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] = 1
            bfs([0, j], islands, visited, new_islands)
        if islands[N-1][j] == 1 and not visited[N-1][j]:
            new_islands[N-1][j] = 1
            bfs([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] = 1
            bfs([i, 0], islands, visited, new_islands)
        if islands[i][M-1] == 1 and not visited[i][M-1]:
            new_islands[i][M-1] = 1
            bfs([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]]:
        return
    visited[cur[0]][cur[1]] = True
    N = 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:
            continue
        if 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 = False
    isSec = False
    for i in range(M):
        if visited[0][i]:
            isFirst = True
            break
    if not isFirst:
        for i in range(N):
            if visited[i][0]:
                isFirst = True
                break
    for i in range(M):
        if visited[N-1][i]:
            isSec = True
            break
    if not isSec:
        for i in range(N):
            if visited[i][M-1]:
                isSec = True
                break   
    if isSec and isFirst:
        return True
    else:
        return False 
if __name__ == '__main__':
    area = 0
    NM = 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]]:
        return
    N = len(islands)
    M = len(islands[0])
    visited[cur[0]][cur[1]] = True
    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:
            continue
        if islands[x][y] < islands[cur[0]][cur[1]]:
            continue
        dfs([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]] = True
    mark[cur[0]][cur[1]] = num
    st = [cur]
    area = 1
    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:
                continue
            if islands[x][y] == 1 and visited[x][y] == False:
                st.append([x, y])
                area += 1
                visited[x][y] = True
                mark[x][y] = num
    return area

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])
    visited = [[False] * M for _ in range(N)]
    mark = [[0] * M for _ in range(N)]
    num = 1
    islandArea = {}
    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] = area
                num += 1
    directions = [[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 = 1
                    pre = []
                    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)