卡码网 99 岛屿数量(深搜版)

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基本就是回溯算法 翻版了

#include <iostream>
#include <vector>using namespace std;int dir[4][2] = {-1, 0, 1, 0, 0, 1, 0, -1};void dfs(vector<vector<int>>& grid, vector<vector<int>>& visited, int i_index, int j_index) {if (!grid[i_index][j_index] || visited[i_index][j_index]) return;visited[i_index][j_index] = 1;for (int i = 0; i < 4; i++) {if (i_index + dir[i][0] >= 0 && i_index + dir[i][0] < grid.size()) {if (j_index + dir[i][1] >= 0 && j_index + dir[i][1] < grid[0].size()) {dfs(grid, visited, i_index + dir[i][0], j_index + dir[i][1]);}}}
}int main() {int n, m;cin >> n >> m;vector<vector<int>> grid(n, vector<int>(m, 0));vector<vector<int>> visited(n, vector<int>(m, 0));int num = 0; for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> grid[i][j];}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (grid[i][j] && !visited[i][j]) {num++;dfs(grid, visited, i, j);}}}cout << num << endl;return 0;
}

卡码网 99 岛屿数量(广搜版)

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类比二叉树层序遍历可以由上面的代码修改得到

#include <iostream>
#include <vector>
#include <queue>
using namespace std;int dir[4][2] = {-1, 0, 1, 0, 0, 1, 0, -1};void bfs(vector<vector<int>>& grid, vector<vector<int>>& visited, int i_index, int j_index) {queue<pair<int, int>> que;que.push(pair<int, int>(i_index, j_index));while (!que.empty()) {// int num = que.size();// while (num--) {i_index = que.front().first;j_index = que.front().second;for (int i = 0; i < 4; i++) {if (i_index + dir[i][0] >= 0 && i_index + dir[i][0] < grid.size()) {if (j_index + dir[i][1] >= 0 && j_index + dir[i][1] < grid[0].size()) {if (grid[i_index+dir[i][0]][j_index+dir[i][1]] && !visited[i_index+dir[i][0]][j_index+dir[i][1]]) {visited[i_index+dir[i][0]][j_index+dir[i][1]] = 1;que.push(pair<int, int>(i_index+dir[i][0],j_index+dir[i][1]));}}}}   que.pop();// }}
}int main() {int n, m;cin >> n >> m;vector<vector<int>> grid(n, vector<int>(m, 0));vector<vector<int>> visited(n, vector<int>(m, 0));int num = 0; for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> grid[i][j];}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (grid[i][j] && !visited[i][j]) {num++;visited[i][j] = 1;bfs(grid, visited, i, j);}}}cout << num << endl;return 0;
}

卡码网 100 岛屿的最大面积

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加了if语句后很神奇地会出问题，改用max，可能判断语句优先级比bfs函数内部执行优先级要高，采用max函数就能解决这个问题

代码如下：

#include <iostream>
#include <vector>
#include <queue>
using namespace std;int dir[4][2] = {-1, 0, 1, 0, 0, 1, 0, -1};int bfs(vector<vector<int>>& grid, vector<vector<int>>& visited, int i_index, int j_index) {int sum = 0;queue<pair<int, int>> que;que.push(pair<int, int>(i_index, j_index));while (!que.empty()) {// int num = que.size();// while (num--) {i_index = que.front().first;j_index = que.front().second;for (int i = 0; i < 4; i++) {if (i_index + dir[i][0] >= 0 && i_index + dir[i][0] < grid.size()) {if (j_index + dir[i][1] >= 0 && j_index + dir[i][1] < grid[0].size()) {if (grid[i_index+dir[i][0]][j_index+dir[i][1]] && !visited[i_index+dir[i][0]][j_index+dir[i][1]]) {visited[i_index+dir[i][0]][j_index+dir[i][1]] = 1;sum++;que.push(pair<int, int>(i_index+dir[i][0],j_index+dir[i][1]));}}}}   que.pop();// }}return sum + 1;
}int main() {int n, m;cin >> n >> m;vector<vector<int>> grid(n, vector<int>(m, 0));vector<vector<int>> visited(n, vector<int>(m, 0));int sum = 0; for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {cin >> grid[i][j];}}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (grid[i][j] && !visited[i][j]) {visited[i][j] = 1;// if (sum < bfs(grid, visited, i, j))sum = max(sum, bfs(grid, visited, i, j));// cout << sum << endl;}}}cout << sum << endl;return 0;
}