506 lines
13 KiB
Markdown
506 lines
13 KiB
Markdown
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<p align="center"><strong><a href="./qita/join.md">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们受益！</strong></p>
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# 102. 沉没孤岛
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[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1174)
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题目描述：
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给定一个由 1（陆地）和 0（水）组成的矩阵，岛屿指的是由水平或垂直方向上相邻的陆地单元格组成的区域，且完全被水域单元格包围。孤岛是那些位于矩阵内部、所有单元格都不接触边缘的岛屿。
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现在你需要将所有孤岛“沉没”，即将孤岛中的所有陆地单元格（1）转变为水域单元格（0）。
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输入描述：
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第一行包含两个整数 N, M，表示矩阵的行数和列数。
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之后 N 行，每行包含 M 个数字，数字为 1 或者 0，表示岛屿的单元格。
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输出描述
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输出将孤岛“沉没”之后的岛屿矩阵。
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输入示例：
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```
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4 5
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1 1 0 0 0
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1 1 0 0 0
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0 0 1 0 0
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0 0 0 1 1
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```
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输出示例：
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```
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1 1 0 0 0
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1 1 0 0 0
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0 0 0 0 0
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0 0 0 1 1
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```
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提示信息：
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将孤岛沉没：
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数据范围：
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1 <= M, N <= 50
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## 思路
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这道题目和[0101.孤岛的总面积](https://kamacoder.com/problempage.php?pid=1173)正好反过来了，[0101.孤岛的总面积](https://kamacoder.com/problempage.php?pid=1173)是求 地图中间的空格数，而本题是要把地图中间的 1 都改成 0 。
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那么两题在思路上也是差不多的。
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思路依然是从地图周边出发，将周边空格相邻的陆地都做上标记，然后在遍历一遍地图，遇到 陆地 且没做过标记的，那么都是地图中间的 陆地 ，全部改成水域就行。
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有的录友可能想，我在定义一个 visited 二维数组，单独标记周边的陆地，然后遍历地图的时候同时对 数组board 和 数组visited 进行判断，决定 陆地是否变成水域。
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这样做其实就有点麻烦了，不用额外定义空间了，标记周边的陆地，可以直接改陆地为其他特殊值作为标记。
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步骤一：深搜或者广搜将地图周边的 1 （陆地）全部改成 2 （特殊标记）
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步骤二：将水域中间 1 （陆地）全部改成 水域（0）
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步骤三：将之前标记的 2 改为 1 （陆地）
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如图：
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整体C++代码如下，以下使用dfs实现，其实遍历方式dfs，bfs都是可以的。
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```CPP
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#include <iostream>
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#include <vector>
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using namespace std;
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int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; // 保存四个方向
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void dfs(vector<vector<int>>& grid, int x, int y) {
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grid[x][y] = 2;
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for (int i = 0; i < 4; i++) { // 向四个方向遍历
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int nextx = x + dir[i][0];
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int nexty = y + dir[i][1];
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// 超过边界
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if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
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// 不符合条件，不继续遍历
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if (grid[nextx][nexty] == 0 || grid[nextx][nexty] == 2) continue;
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dfs (grid, nextx, nexty);
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}
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return;
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}
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int main() {
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int n, m;
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cin >> n >> m;
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vector<vector<int>> grid(n, vector<int>(m, 0));
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++) {
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cin >> grid[i][j];
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}
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}
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// 步骤一：
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// 从左侧边，和右侧边 向中间遍历
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for (int i = 0; i < n; i++) {
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if (grid[i][0] == 1) dfs(grid, i, 0);
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if (grid[i][m - 1] == 1) dfs(grid, i, m - 1);
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}
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// 从上边和下边 向中间遍历
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for (int j = 0; j < m; j++) {
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if (grid[0][j] == 1) dfs(grid, 0, j);
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if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
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}
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// 步骤二、步骤三
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++) {
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if (grid[i][j] == 1) grid[i][j] = 0;
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if (grid[i][j] == 2) grid[i][j] = 1;
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}
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}
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++) {
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cout << grid[i][j] << " ";
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}
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cout << endl;
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}
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}
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```
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## 其他语言版本
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### Java
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```JAVA
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import java.util.Scanner;
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public class Main {
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static int[][] dir = { {-1, 0}, {0, -1}, {1, 0}, {0, 1} }; // 保存四个方向
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public static void dfs(int[][] grid, int x, int y) {
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grid[x][y] = 2;
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for (int[] d : dir) {
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int nextX = x + d[0];
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int nextY = y + d[1];
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// 超过边界
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if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length) continue;
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// 不符合条件，不继续遍历
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if (grid[nextX][nextY] == 0 || grid[nextX][nextY] == 2) continue;
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dfs(grid, nextX, nextY);
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}
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}
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public static void main(String[] args) {
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Scanner scanner = new Scanner(System.in);
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int n = scanner.nextInt();
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int m = scanner.nextInt();
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int[][] grid = new int[n][m];
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++) {
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grid[i][j] = scanner.nextInt();
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}
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}
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// 步骤一：
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// 从左侧边，和右侧边 向中间遍历
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for (int i = 0; i < n; i++) {
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if (grid[i][0] == 1) dfs(grid, i, 0);
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if (grid[i][m - 1] == 1) dfs(grid, i, m - 1);
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}
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// 从上边和下边 向中间遍历
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for (int j = 0; j < m; j++) {
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if (grid[0][j] == 1) dfs(grid, 0, j);
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if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
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}
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// 步骤二、步骤三
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++) {
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if (grid[i][j] == 1) grid[i][j] = 0;
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if (grid[i][j] == 2) grid[i][j] = 1;
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}
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}
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++) {
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System.out.print(grid[i][j] + " ");
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}
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System.out.println();
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}
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scanner.close();
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}
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}
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```
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### Python
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```python
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def dfs(grid, x, y):
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grid[x][y] = 2
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directions = [(-1, 0), (0, -1), (1, 0), (0, 1)] # 四个方向
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for dx, dy in directions:
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nextx, nexty = x + dx, y + dy
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# 超过边界
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if nextx < 0 or nextx >= len(grid) or nexty < 0 or nexty >= len(grid[0]):
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continue
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# 不符合条件，不继续遍历
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if grid[nextx][nexty] == 0 or grid[nextx][nexty] == 2:
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continue
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dfs(grid, nextx, nexty)
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def main():
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n, m = map(int, input().split())
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grid = [[int(x) for x in input().split()] for _ in range(n)]
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# 步骤一：
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# 从左侧边，和右侧边 向中间遍历
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for i in range(n):
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if grid[i][0] == 1:
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dfs(grid, i, 0)
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if grid[i][m - 1] == 1:
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dfs(grid, i, m - 1)
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# 从上边和下边 向中间遍历
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for j in range(m):
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if grid[0][j] == 1:
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dfs(grid, 0, j)
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if grid[n - 1][j] == 1:
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dfs(grid, n - 1, j)
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# 步骤二、步骤三
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for i in range(n):
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for j in range(m):
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if grid[i][j] == 1:
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grid[i][j] = 0
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if grid[i][j] == 2:
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grid[i][j] = 1
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# 打印结果
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for row in grid:
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print(' '.join(map(str, row)))
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if __name__ == "__main__":
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main()
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```
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广搜版
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```Python
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from collections import deque
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n, m = list(map(int, input().split()))
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g = []
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for _ in range(n):
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row = list(map(int,input().split()))
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g.append(row)
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directions = [(1,0),(-1,0),(0,1),(0,-1)]
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count = 0
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def bfs(r,c,mode):
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global count
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q = deque()
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q.append((r,c))
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count += 1
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while q:
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r, c = q.popleft()
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if mode:
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g[r][c] = 2
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for di in directions:
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next_r = r + di[0]
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next_c = c + di[1]
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if next_c < 0 or next_c >= m or next_r < 0 or next_r >= n:
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continue
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if g[next_r][next_c] == 1:
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q.append((next_r,next_c))
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if mode:
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g[r][c] = 2
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count += 1
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for i in range(n):
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if g[i][0] == 1: bfs(i,0,True)
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if g[i][m-1] == 1: bfs(i, m-1,True)
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for j in range(m):
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if g[0][j] == 1: bfs(0,j,1)
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if g[n-1][j] == 1: bfs(n-1,j,1)
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for i in range(n):
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for j in range(m):
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if g[i][j] == 2:
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g[i][j] = 1
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else:
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g[i][j] = 0
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for row in g:
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print(" ".join(map(str, row)))
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```
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### Go
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### Rust
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### Javascript
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#### 深搜版
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```javascript
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const r1 = require('readline').createInterface({ input: process.stdin });
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// 创建readline接口
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let iter = r1[Symbol.asyncIterator]();
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// 创建异步迭代器
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const readline = async () => (await iter.next()).value;
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let graph // 地图
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let N, M // 地图大小
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const dir = [[0, 1], [1, 0], [0, -1], [-1, 0]] //方向
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// 读取输入，初始化地图
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const initGraph = async () => {
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let line = await readline();
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[N, M] = line.split(' ').map(Number);
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graph = new Array(N).fill(0).map(() => new Array(M).fill(0))
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for (let i = 0; i < N; i++) {
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line = await readline()
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line = line.split(' ').map(Number)
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for (let j = 0; j < M; j++) {
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graph[i][j] = line[j]
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}
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}
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}
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/**
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* @description: 从（x，y）开始深度优先遍历地图
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* @param {*} graph 地图
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* @param {*} x 开始搜索节点的下标
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* @param {*} y 开始搜索节点的下标
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* @return {*}
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*/
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const dfs = (graph, x, y) => {
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if (graph[x][y] !== 1) return
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graph[x][y] = 2 // 标记为非孤岛陆地
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for (let i = 0; i < 4; i++) {
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let nextx = x + dir[i][0]
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let nexty = y + dir[i][1]
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if (nextx < 0 || nextx >= N || nexty < 0 || nexty >= M) continue
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dfs(graph, nextx, nexty)
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}
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}
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(async function () {
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// 读取输入，初始化地图
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await initGraph()
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// 遍历地图左右两边
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for (let i = 0; i < N; i++) {
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if (graph[i][0] === 1) dfs(graph, i, 0)
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if (graph[i][M - 1] === 1) dfs(graph, i, M - 1)
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}
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// 遍历地图上下两边
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for (let j = 0; j < M; j++) {
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if (graph[0][j] === 1) dfs(graph, 0, j)
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if (graph[N - 1][j] === 1) dfs(graph, N - 1, j)
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}
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// 遍历地图，将孤岛沉没，即将陆地1标记为0；将非孤岛陆地2标记为1
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for (let i = 0; i < N; i++) {
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for (let j = 0; j < M; j++) {
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if (graph[i][j] === 1) graph[i][j] = 0
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else if (graph[i][j] === 2) graph[i][j] = 1
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}
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console.log(graph[i].join(' '));
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}
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})()
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```
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#### 广搜版
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```javascript
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const r1 = require('readline').createInterface({ input: process.stdin });
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// 创建readline接口
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let iter = r1[Symbol.asyncIterator]();
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// 创建异步迭代器
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const readline = async () => (await iter.next()).value;
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let graph // 地图
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let N, M // 地图大小
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const dir = [[0, 1], [1, 0], [0, -1], [-1, 0]] //方向
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// 读取输入，初始化地图
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const initGraph = async () => {
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let line = await readline();
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[N, M] = line.split(' ').map(Number);
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graph = new Array(N).fill(0).map(() => new Array(M).fill(0))
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for (let i = 0; i < N; i++) {
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line = await readline()
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line = line.split(' ').map(Number)
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for (let j = 0; j < M; j++) {
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graph[i][j] = line[j]
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}
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}
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}
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/**
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* @description: 从（x，y）开始广度优先遍历地图
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* @param {*} graph 地图
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* @param {*} x 开始搜索节点的下标
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* @param {*} y 开始搜索节点的下标
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* @return {*}
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*/
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const bfs = (graph, x, y) => {
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let queue = []
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queue.push([x, y])
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graph[x][y] = 2 // 标记为非孤岛陆地
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while (queue.length) {
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let [xx, yy] = queue.shift()
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for (let i = 0; i < 4; i++) {
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let nextx = xx + dir[i][0]
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let nexty = yy + dir[i][1]
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if (nextx < 0 || nextx >= N || nexty < 0 || nexty >= M) continue
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if (graph[nextx][nexty] === 1) bfs(graph, nextx, nexty)
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}
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}
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}
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(async function () {
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// 读取输入，初始化地图
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await initGraph()
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// 遍历地图左右两边
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for (let i = 0; i < N; i++) {
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if (graph[i][0] === 1) bfs(graph, i, 0)
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if (graph[i][M - 1] === 1) bfs(graph, i, M - 1)
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}
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// 遍历地图上下两边
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for (let j = 0; j < M; j++) {
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if (graph[0][j] === 1) bfs(graph, 0, j)
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if (graph[N - 1][j] === 1) bfs(graph, N - 1, j)
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}
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// 遍历地图，将孤岛沉没，即将陆地1标记为0；将非孤岛陆地2标记为1
|
||
for (let i = 0; i < N; i++) {
|
||
for (let j = 0; j < M; j++) {
|
||
if (graph[i][j] === 1) graph[i][j] = 0
|
||
else if (graph[i][j] === 2) graph[i][j] = 1
|
||
}
|
||
console.log(graph[i].join(' '));
|
||
}
|
||
})()
|
||
```
|
||
|
||
|
||
|
||
### TypeScript
|
||
|
||
### PhP
|
||
|
||
### Swift
|
||
|
||
### Scala
|
||
|
||
### C#
|
||
|
||
### Dart
|
||
|
||
### C
|
||
|