101 孤岛的总面积
题目链接
思路
将每个岛的最外沿先都变成陆地，再遍历即可
文章详解·
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] = 0;
for(int i = 0; i < 4; i++){
int next_x = x + dir[i];
int next_y = y + dir[i];
if(next_x >= 0 && next_x < grid.size() && next_y >= 0 && next_y < grid[0].size())
{
if(grid[next_x][next_y] == 1){
dfs(grid,next_x,next_y);
}
}
}
}
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];
}
}
for(int i = 0; i < n; i++)
{
if(grid[i][0] == 1){
dfs(grid,0,i);
}
if(grid[i][m-1] == 1){
dfs(grid,m-1,i);
}
}
for(int j = 0; j < m; j++)
{
if(grid[0][j] == 1){
dfs(grid,j,0);
}
if(grid[n-1][j] == 1){
dfs(grid,j,n-1);
}
}
int count = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) count++;
}
}
cout << count << endl;
}102 沉没孤岛
题目链接
思路
和上题类似，将输入复制一份，先遍历靠边的海岸，根据第一幅图找到的孤岛，在第二幅图中沉没即可
文章详解
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] = 0;
for(int i = 0; i < 4; i++){
int next_x = x + dir[i][0];
int next_y = y + dir[i][1];
if(next_x >= 0 && next_x < grid.size() && next_y >= 0 && next_y < grid[0].size())
{
if(grid[next_x][next_y] == 1){
dfs(grid,next_x,next_y);
}
}
}
}
int main() {
int n, m;
cin >> n >> m;
vector<vector<int>> grid(n, vector<int>(m, 0));
vector<vector<int>> grid_copy(n, vector<int>(m, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> grid[i][j];
grid_copy[i][j] = grid[i][j];
}
}
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);
}
}
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);
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1){
grid_copy[i][j] = 0;
}
cout << grid_copy[i][j] << " ";
}
cout << endl;
}
}103 水流问题
题目链接
思路
从边界出发，逆流而上，第一组边界和第二组边界都标记的点就是结果集中的一个。
文章详解
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
思路
第一次遍历记录信息，第二次遍历所有海洋格子计算相邻陆地面积。精妙之处在于：如何确认相邻陆地属于哪一块陆地？不好记录整体位置信息，不妨化整为零：我不记录你，你向我报告你属于哪一块地。
文章详解
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,化整为零
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;
}