目录
101.孤岛的总面积
-
题目链接：101. 孤岛的总面积
-
文章讲解：代码随想录
图论法
- 代码一：深搜
cpp
#include <iostream>
#include <vector>
using namespace std;
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; // 保存四个方向
int count; // 统计符合题目要求的陆地空格数量
void dfs(vector<vector<int>>& grid, int x, int y) {
grid[x][y] = 0;
count++;
for (int i = 0; i < 4; i++) { // 向四个方向遍历
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
// 超过边界
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
// 不符合条件，不继续遍历
if (grid[nextx][nexty] == 0) continue;
dfs (grid, nextx, nexty);
}
return;
}
int main() {
int n, m;
cin >> n >> m;
vector<vector<int>> grid(n, vector<int>(m, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> grid[i][j];
}
}
// 从左侧边，和右侧边 向中间遍历
for (int i = 0; i < n; i++) {
if (grid[i][0] == 1) dfs(grid, i, 0);
if (grid[i][m - 1] == 1) dfs(grid, i, m - 1);
}
// 从上边和下边 向中间遍历
for (int j = 0; j < m; j++) {
if (grid[0][j] == 1) dfs(grid, 0, j);
if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
}
count = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) dfs(grid, i, j);
}
}
cout << count << endl;
}- 代码二：广搜
cpp
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int count = 0;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void bfs(vector<vector<int>>& grid, int x, int y) {
queue<pair<int, int>> que;
que.push({x, y});
grid[x][y] = 0; // 只要加入队列，立刻标记
count++;
while(!que.empty()) {
pair<int ,int> cur = que.front(); que.pop();
int curx = cur.first;
int cury = cur.second;
for (int i = 0; i < 4; i++) {
int nextx = curx + dir[i][0];
int nexty = cury + dir[i][1];
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue; // 越界了，直接跳过
if (grid[nextx][nexty] == 1) {
que.push({nextx, nexty});
count++;
grid[nextx][nexty] = 0; // 只要加入队列立刻标记
}
}
}
}
int main() {
int n, m;
cin >> n >> m;
vector<vector<int>> grid(n, vector<int>(m, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> grid[i][j];
}
}
// 从左侧边，和右侧边 向中间遍历
for (int i = 0; i < n; i++) {
if (grid[i][0] == 1) bfs(grid, i, 0);
if (grid[i][m - 1] == 1) bfs(grid, i, m - 1);
}
// 从上边和下边 向中间遍历
for (int j = 0; j < m; j++) {
if (grid[0][j] == 1) bfs(grid, 0, j);
if (grid[n - 1][j] == 1) bfs(grid, n - 1, j);
}
count = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) bfs(grid, i, j);
}
}
cout << count << endl;
}102.沉没孤岛
图论法
- 代码一： dfs
cpp
#include <iostream>
#include <vector>
using namespace std;
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1}; // 保存四个方向
void dfs(vector<vector<int>>& grid, int x, int y) {
grid[x][y] = 2;
for (int i = 0; i < 4; i++) { // 向四个方向遍历
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
// 超过边界
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
// 不符合条件，不继续遍历
if (grid[nextx][nexty] == 0 || grid[nextx][nexty] == 2) continue;
dfs (grid, nextx, nexty);
}
return;
}
int main() {
int n, m;
cin >> n >> m;
vector<vector<int>> grid(n, vector<int>(m, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> grid[i][j];
}
}
// 步骤一：
// 从左侧边，和右侧边 向中间遍历
for (int i = 0; i < n; i++) {
if (grid[i][0] == 1) dfs(grid, i, 0);
if (grid[i][m - 1] == 1) dfs(grid, i, m - 1);
}
// 从上边和下边 向中间遍历
for (int j = 0; j < m; j++) {
if (grid[0][j] == 1) dfs(grid, 0, j);
if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
}
// 步骤二、步骤三
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) grid[i][j] = 0;
if (grid[i][j] == 2) grid[i][j] = 1;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cout << grid[i][j] << " ";
}
cout << endl;
}
}103.水流问题
图论法
- 代码一： dfs
cpp
#include <iostream>
#include <vector>
using namespace std;
int n, m;
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1};
void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y) {
if (visited[x][y]) return;
visited[x][y] = true;
for (int i = 0; i < 4; i++) {
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if (nextx < 0 || nextx >= n || nexty < 0 || nexty >= m) continue;
if (grid[x][y] > grid[nextx][nexty]) continue; // 注意：这里是从低向高遍历
dfs (grid, visited, nextx, nexty);
}
return;
}
int main() {
cin >> n >> m;
vector<vector<int>> grid(n, vector<int>(m, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> grid[i][j];
}
}
// 标记从第一组边界上的节点出发，可以遍历的节点
vector<vector<bool>> firstBorder(n, vector<bool>(m, false));
// 标记从第一组边界上的节点出发，可以遍历的节点
vector<vector<bool>> secondBorder(n, vector<bool>(m, false));
// 从最上和最下行的节点出发，向高处遍历
for (int i = 0; i < n; i++) {
dfs (grid, firstBorder, i, 0); // 遍历最左列，接触第一组边界
dfs (grid, secondBorder, i, m - 1); // 遍历最右列，接触第二组边界
}
// 从最左和最右列的节点出发，向高处遍历
for (int j = 0; j < m; j++) {
dfs (grid, firstBorder, 0, j); // 遍历最上行，接触第一组边界
dfs (grid, secondBorder, n - 1, j); // 遍历最下行，接触第二组边界
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// 如果这个节点，从第一组边界和第二组边界出发都遍历过，就是结果
if (firstBorder[i][j] && secondBorder[i][j]) cout << i << " " << j << endl;;
}
}
}104.建造最大岛屿
-
题目链接：104. 建造最大岛屿
-
文章讲解：代码随想录
图论法
- 代码一：深搜
cpp
#include <iostream>
#include <vector>
#include <unordered_set>
#include <unordered_map>
using namespace std;
int n, m;
int count;
int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1}; // 四个方向
void dfs(vector<vector<int>>& grid, vector<vector<bool>>& visited, int x, int y, int mark) {
if (visited[x][y] || grid[x][y] == 0) return; // 终止条件：访问过的节点 或者 遇到海水
visited[x][y] = true; // 标记访问过
grid[x][y] = mark; // 给陆地标记新标签
count++;
for (int i = 0; i < 4; i++) {
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
if (nextx < 0 || nextx >= n || nexty < 0 || nexty >= m) continue; // 越界了，直接跳过
dfs(grid, visited, nextx, nexty, mark);
}
}
int main() {
cin >> n >> m;
vector<vector<int>> grid(n, vector<int>(m, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> grid[i][j];
}
}
vector<vector<bool>> visited(n, vector<bool>(m, false)); // 标记访问过的点
unordered_map<int ,int> gridNum;
int mark = 2; // 记录每个岛屿的编号
bool isAllGrid = true; // 标记是否整个地图都是陆地
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 0) isAllGrid = false;
if (!visited[i][j] && grid[i][j] == 1) {
count = 0;
dfs(grid, visited, i, j, mark); // 将与其链接的陆地都标记上 true
gridNum[mark] = count; // 记录每一个岛屿的面积
mark++; // 记录下一个岛屿编号
}
}
}
if (isAllGrid) {
cout << n * m << endl; // 如果都是陆地，返回全面积
return 0; // 结束程序
}
// 以下逻辑是根据添加陆地的位置，计算周边岛屿面积之和
int result = 0; // 记录最后结果
unordered_set<int> visitedGrid; // 标记访问过的岛屿
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
count = 1; // 记录连接之后的岛屿数量
visitedGrid.clear(); // 每次使用时，清空
if (grid[i][j] == 0) {
for (int k = 0; k < 4; k++) {
int neari = i + dir[k][1]; // 计算相邻坐标
int nearj = j + dir[k][0];
if (neari < 0 || neari >= n || nearj < 0 || nearj >= m) continue;
if (visitedGrid.count(grid[neari][nearj])) continue; // 添加过的岛屿不要重复添加
// 把相邻四面的岛屿数量加起来
count += gridNum[grid[neari][nearj]];
visitedGrid.insert(grid[neari][nearj]); // 标记该岛屿已经添加过
}
}
result = max(result, count);
}
}
cout << result << endl;
}