463 lines
12 KiB
Markdown
463 lines
12 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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# 99. 岛屿数量
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[卡码网题目链接（ACM模式）](https://kamacoder.com/problempage.php?pid=1171)
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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，你需要计算岛屿的数量。岛屿由水平方向或垂直方向上相邻的陆地连接而成，并且四周都是水域。你可以假设矩阵外均被水包围。
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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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后续 N 行，每行包含 M 个数字，数字为 1 或者 0。
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输出描述：
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输出一个整数，表示岛屿的数量。如果不存在岛屿，则输出 0。
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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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3
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提示信息
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根据测试案例中所展示，岛屿数量共有 3 个，所以输出 3。
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数据范围：
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* 1 <= N, M <= 50
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## 思路
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注意题目中每座岛屿只能由**水平方向和/或竖直方向上**相邻的陆地连接形成。
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也就是说斜角度链接是不算了， 例如示例二，是三个岛屿，如图：
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这道题题目是 DFS，BFS，并查集，基础题目。
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本题思路:遇到一个没有遍历过的节点陆地，计数器就加一，然后把该节点陆地所能遍历到的陆地都标记上。
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再遇到标记过的陆地节点和海洋节点的时候直接跳过。 这样计数器就是最终岛屿的数量。
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那么如果把节点陆地所能遍历到的陆地都标记上呢，就可以使用 DFS，BFS或者并查集。
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### 广度优先搜索
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如果不熟悉广搜，建议先看 [广搜理论基础](./图论广搜理论基础.md)。
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不少同学用广搜做这道题目的时候，超时了。 这里有一个广搜中很重要的细节：
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根本原因是**只要 加入队列就代表 走过，就需要标记，而不是从队列拿出来的时候再去标记走过**。
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很多同学可能感觉这有区别吗？
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如果从队列拿出节点，再去标记这个节点走过，就会发生下图所示的结果，会导致很多节点重复加入队列。
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超时写法 （从队列中取出节点再标记，注意代码注释的地方）
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```CPP
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int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
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void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
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queue<pair<int, int>> que;
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que.push({x, y});
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while(!que.empty()) {
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pair<int ,int> cur = que.front(); que.pop();
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int curx = cur.first;
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int cury = cur.second;
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visited[curx][cury] = true; // 从队列中取出在标记走过
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for (int i = 0; i < 4; i++) {
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int nextx = curx + dir[i][0];
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int nexty = cury + dir[i][1];
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if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
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if (!visited[nextx][nexty] && grid[nextx][nexty] == '1') {
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que.push({nextx, nexty});
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}
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}
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}
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}
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```
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加入队列 就代表走过，立刻标记，正确写法： （注意代码注释的地方）
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```CPP
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int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
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void bfs(vector<vector<char>>& grid, vector<vector<bool>>& visited, int x, int y) {
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queue<pair<int, int>> que;
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que.push({x, y});
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visited[x][y] = true; // 只要加入队列，立刻标记
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while(!que.empty()) {
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pair<int ,int> cur = que.front(); que.pop();
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int curx = cur.first;
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int cury = cur.second;
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for (int i = 0; i < 4; i++) {
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int nextx = curx + dir[i][0];
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int nexty = cury + dir[i][1];
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if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
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if (!visited[nextx][nexty] && grid[nextx][nexty] == '1') {
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que.push({nextx, nexty});
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visited[nextx][nexty] = true; // 只要加入队列立刻标记
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}
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}
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}
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}
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```
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以上两个版本其实，其实只有细微区别，就是 `visited[x][y] = true;` 放在的地方，这取决于我们对 代码中队列的定义，队列中的节点就表示已经走过的节点。 **所以只要加入队列，立即标记该节点走过**。
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本题完整广搜代码：
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```CPP
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#include <iostream>
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#include <vector>
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#include <queue>
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using namespace std;
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int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
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void bfs(const vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
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queue<pair<int, int>> que;
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que.push({x, y});
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visited[x][y] = true; // 只要加入队列，立刻标记
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while(!que.empty()) {
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pair<int ,int> cur = que.front(); que.pop();
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int curx = cur.first;
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int cury = cur.second;
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for (int i = 0; i < 4; i++) {
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int nextx = curx + dir[i][0];
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int nexty = cury + dir[i][1];
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if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
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if (!visited[nextx][nexty] && grid[nextx][nexty] == 1) {
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que.push({nextx, nexty});
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visited[nextx][nexty] = true; // 只要加入队列立刻标记
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}
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}
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}
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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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vector<vector<bool>> visited(n, vector<bool>(m, false));
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int result = 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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if (!visited[i][j] && grid[i][j] == 1) {
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result++; // 遇到没访问过的陆地，+1
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bfs(grid, visited, i, j); // 将与其链接的陆地都标记上 true
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}
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}
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}
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cout << result << endl;
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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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```
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### Python
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```python
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from collections import deque
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directions = [[0, 1], [1, 0], [0, -1], [-1, 0]]
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def bfs(grid, visited, x, y):
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que = deque([])
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que.append([x,y])
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while que:
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cur_x, cur_y = que.popleft()
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for i, j in directions:
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next_x = cur_x + i
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next_y = cur_y + j
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if next_y < 0 or next_x < 0 or next_x >= len(grid) or next_y >= len(grid[0]):
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continue
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if not visited[next_x][next_y] and grid[next_x][next_y] == 1:
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visited[next_x][next_y] = True
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que.append([next_x, next_y])
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def main():
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n, m = map(int, input().split())
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grid = []
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for i in range(n):
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grid.append(list(map(int, input().split())))
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visited = [[False] * m for _ in range(n)]
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res = 0
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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 and not visited[i][j]:
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res += 1
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bfs(grid, visited, i, j)
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print(res)
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if __name__ == "__main__":
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main()
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```
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### Go
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```go
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package main
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import (
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"bufio"
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"fmt"
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"os"
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"strconv"
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"strings"
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)
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var dir = [4][2]int{{0, 1}, {1, 0}, {-1, 0}, {0, -1}} // 四个方向
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func dfs(grid [][]int, visited [][]bool, x, y int) {
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for i := 0; i < 4; i++ {
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nextx := x + dir[i][0]
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nexty := y + dir[i][1]
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if nextx < 0 || nextx >= len(grid) || nexty < 0 || nexty >= len(grid[0]) {
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continue // 越界了，直接跳过
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}
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if !visited[nextx][nexty] && grid[nextx][nexty] == 1 { // 没有访问过的 同时 是陆地的
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visited[nextx][nexty] = true
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dfs(grid, visited, nextx, nexty)
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}
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}
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}
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func main() {
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reader := bufio.NewReader(os.Stdin)
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var n, m int
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fmt.Scanf("%d %d", &n, &m)
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grid := make([][]int, n)
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for i := 0; i < n; i++ {
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grid[i] = make([]int, m)
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line, _ := reader.ReadString('\n')
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line = strings.TrimSpace(line)
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elements := strings.Split(line, " ")
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for j := 0; j < m; j++ {
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grid[i][j], _ = strconv.Atoi(elements[j])
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}
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}
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visited := make([][]bool, n)
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for i := 0; i < n; i++ {
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visited[i] = make([]bool, m)
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}
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result := 0
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for i := 0; i < n; i++ {
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for j := 0; j < m; j++ {
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if !visited[i][j] && grid[i][j] == 1 {
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visited[i][j] = true
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result++ // 遇到没访问过的陆地，+1
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dfs(grid, visited, i, j) // 将与其链接的陆地都标记上 true
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}
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}
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}
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fmt.Println(result)
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}
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```
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### Rust
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### Javascript
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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
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let N, M
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let visited
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let result = 0
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const dir = [[0, 1], [1, 0], [0, -1], [-1, 0]]
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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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visited = new Array(N).fill(false).map(() => new Array(M).fill(false))
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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
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* @param {*} graph 地图
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* @param {*} visited 访问过的节点
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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, visited, x, y) => {
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let queue = []
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queue.push([x, y])
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visited[x][y] = true //只要加入队列就立刻标记为访问过
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while (queue.length) {
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let [x, y] = queue.shift()
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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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if(!visited[nextx][nexty] && graph[nextx][nexty] === 1){
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queue.push([nextx, nexty])
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visited[nextx][nexty] = true
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}
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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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// 统计岛屿数
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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 (!visited[i][j] && graph[i][j] === 1) {
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// 遇到没访问过的陆地，+1
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result++
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// 广度优先遍历，将相邻陆地标记为已访问
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bfs(graph, visited, i, j)
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}
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}
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}
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console.log(result);
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})()
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```
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### TypeScript
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### PhP
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```PHP
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<?php
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function dfs($grid, &$visited, $x, $y) {
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$dir = [[0, 1], [1, 0], [-1, 0], [0, -1]]; // 四个方向
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foreach ($dir as $d) {
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$nextx = $x + $d[0];
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$nexty = $y + $d[1];
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if ($nextx < 0 || $nextx >= count($grid) || $nexty < 0 || $nexty >= count($grid[0])) {
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continue; // 越界了，直接跳过
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}
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if (!$visited[$nextx][$nexty] && $grid[$nextx][$nexty] == 1) { // 没有访问过的 同时 是陆地的
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$visited[$nextx][$nexty] = true;
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dfs($grid, $visited, $nextx, $nexty);
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}
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}
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}
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function main() {
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fscanf(STDIN, "%d %d", $n, $m);
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$grid = [];
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for ($i = 0; $i < $n; $i++) {
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$grid[$i] = array_map('intval', explode(' ', trim(fgets(STDIN))));
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}
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$visited = array_fill(0, $n, array_fill(0, $m, false));
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$result = 0;
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for ($i = 0; $i < $n; $i++) {
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for ($j = 0; $j < $m; $j++) {
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if (!$visited[$i][$j] && $grid[$i][$j] == 1) {
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$visited[$i][$j] = true;
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$result++; // 遇到没访问过的陆地，+1
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dfs($grid, $visited, $i, $j); // 将与其链接的陆地都标记上 true
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}
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}
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}
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echo $result . PHP_EOL;
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}
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main();
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?>
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```
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### Swift
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### Scala
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### C#
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|
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|
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### Dart
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|
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### C
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|