【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] = 0
    for 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]):
            continue
        if 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 = 0
    for i in range(n):
        for j in range(m):
            if graph[i][j] == 1:
                result += 1
    print(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] = 0
        for 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]):
                continue
            if 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 = 0
    for i in range(n):
        for j in range(m):
            if graph[i][j] == 1:
                result += 1
    print(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] = 2
    for 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]):
            continue
        if 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] = 0
    for i in range(n):
        for j in range(m):
            if graph[i][j] == 2:
                graph[i][j] = 1
    for 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] = 2
        for 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]):
                continue
            if 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] = 0
    for i in range(n):
        for j in range(m):
            if graph[i][j] == 2:
                graph[i][j] = 1
    for 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] = True
    for 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]):
            continue
        if graph[nextx][nexty] <= graph[x][y] and visited[nextx][nexty] == False:
            if nextx == 0 or nexty == 0:
                local_contact[x][y][0] = True
            if nextx == n - 1 or nexty == m - 1:
                local_contact[x][y][1] = True
            visited[nextx][nexty] = True
            dfs(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] = True
    for 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]):
            continue
        if graph[nextx][nexty] >= graph[x][y] and visited[nextx][nexty] == False:
            #visited[nextx][nexty] = True
            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())))
    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