代码随想录训练营Day 66|卡码网101.孤岛的总面积、102.沉没孤岛、103.水流问题、104.建造最大岛屿

1.孤岛的总面积

101. 孤岛的总面积 | 代码随想录

代码：(bfs广搜)

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int dir[4][2] = {1,0,0,1,-1,0,0,-1};
int count;
void bfs(vector<vector<int>>&grid,int x,int y){queue<pair<int,int>> que;que.push({x,y});grid[x][y] = 0; // 这里用grid的元素变为0来进行标记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;}if(grid[nextx][nexty] == 1){que.push({nextx,nexty});grid[nextx][nexty] = 0;count++;}}}
}
int main(){// 输入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];}}// 处理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);}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 = 1; i < n - 1; i++){for(int j = 1; j < m - 1; j++){if(grid[i][j] == 1) bfs(grid,i,j);}}// 输出cout << count << endl;
}

思路：基础题型，这次直接在grid表上进行标记。先把靠近陆地的岛屿都标为0。将用于统计面积的变量的count置为0。接下来就可以排除临近陆地的岛屿，只统计孤岛的数量了。 

易错点：我又犯错了。先是忘记写命名空间了，然后又是在if判断里二维数组我只写了一个下标，最后是找边界找错了，我的循环变量i是在变的，我只要固定另一个下标为临界值就好了。不知道为什么我把i放到了边界的运算里（。。脑子糊了）

代码：（dfs 深搜）

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int dir[4][2] = {1,0,0,1,-1,0,0,-1};
int count;
void bfs(vector<vector<int>>&grid,int x,int y){if(grid[x][y] == 0){return;}grid[x][y] = 0;count++;for(int i = 0; i < 4; i++){int nextx = x + dir[i][0];int nexty = y + dir[i][1];if(nextx < 0||nextx >= grid.size()||nexty < 0||nexty >= grid[0].size()){continue;}bfs(grid,nextx,nexty);}
}
int main(){// 输入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];}}// 处理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);}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 = 1; i < n - 1; i++){for(int j = 1; j < m - 1; j++){if(grid[i][j] == 1) bfs(grid,i,j);}}// 输出cout << count << endl;
}

 思路：我的博客快做成个人错误大赏了。。。

这次的错误是 在进行下标是否非法的判断的时候，我又写错了。一定要注意边界条件啊，以后检查的时候就应该找这种判断语句好好看看。

2.沉没孤岛 

102. 沉没孤岛 | 代码随想录

代码：（深搜dfs）

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int dir[4][2] = {1,0,0,1,-1,0,0,-1};
int count;
void bfs(vector<vector<int>>&grid,int x,int y){if(grid[x][y] == 2||grid[x][y] == 0){return;}grid[x][y] = 2;for(int i = 0; i < 4; i++){int nextx = x + dir[i][0];int nexty = y + dir[i][1];if(nextx < 0||nextx >= grid.size()||nexty < 0||nexty >= grid[0].size()){continue;}bfs(grid,nextx,nexty);}
}
int main(){// 输入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];}}// 处理// 将临近陆地的标为2 和 孤岛区分开vector<vector<bool>> visited(n,vector<bool>(m,false));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);}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);}// 将孤岛沉没 将临近陆地的单元还原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;}}// 输出for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){cout << grid[i][j] << " ";}cout << endl;}
}

思路：这里比较巧妙的就是因为临近陆地的岛屿比较好判断，所以将其标记为2，只要与孤岛区别开就好了。接下来，就是把1变为0，把2变为1，输出就好了 

3.水流问题

103. 水流问题 | 代码随想录

代码： (深搜dfs)

#include <iostream>
#include <vector>
using namespace std;
int dir[4][2] = {0,1,1,0,0,-1,-1,0};
void dfs(const 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][4];if(nextx < 0||nextx >= grid.size()||nexty < 0||nexty >= grid[0].size()){continue;}// 必须从低到高if(grid[nextx][nexty] < grid[x][y]) continue;dfs(grid,visited,nextx,nexty);}return;
}
int main(){//输入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];}}// 处理vector<vector<bool>> firstBorder(n,vector<bool>(m,false));vector<vector<bool>> secondBorder(n,vector<bool>(m,false));for(int i = 0; i < n; i++){dfs(grid,firstBorder,i,0);dfs(grid,secondBorderBorder,i,m - 1);}for(int j = 0; j < m; j++){dfs(grid,firstBorder,0,j);dfs(grid,secondBorder,n - 1,j);}// 输出for(int i = 0; i < n; i++){for(int j = 0; j < m; j++){if(firstBorder[i][j]&&secondBorder[i][j]){cout << i << j << endl;}}}
}

 思路：这道题的优化很巧妙，因为从边界出发好判断，所以直接从边界出发，把边界可以到达的地方而且是递增的单元进行标记。上边界和下边界各用一套标记，当一个单元被标记两次时，就说明它时满足题意的。

易错点：我自己写了半天，忘记写最重要的 if(grid[nextx][nexty] < grid[x][y]) continue;了。

4.建造最大岛屿 

 104.建造最大岛屿 | 代码随想录

代码： 

#include <iostream>
#include <vector>
#include <unordered_set>
#include <unordered_map>
using namespace std;
int count;
int dir[4][2] = {0,1,1,0,0,-1,-1,0};
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;visited[x][y] = true; // 标记已访问grid[x][y] = mark; // 每个岛屿都有对应的标签count++;for(int i = 0; i < 4; i++){int nextx = x + dir[i][0];int nexty = y + dir[i][1];if(nextx < 0||nextx >= grid.size()||nexty < 0||nexty >= grid[0].size()){continue;}dfs(grid,visited,nextx,nexty,mark);}return;
}
int main(){// 输入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];}}// 处理vector<vector<bool>> visited(n,vector<bool>(m,false));unordered_map<int,int> gridNum;int mark = 2; // 记录岛屿的编号bool isAllGrid = true; // 记录整个地图是否都是陆地(如果是，就没法添加陆地了)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);gridNum[mark] = count; // 记录每一个岛屿的面积mark++;}}}if(isAllGrid == true) {cout << n*m << endl;return 0;}// 根据所填陆地，计算周边岛屿面积之和int result = 0;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; // 记录连接后的岛屿数量visitedGrid.clear();if(grid[i][j] == 0){for(int k = 0; k < 4; k++){int nexti = i + dir[k][0];int nextj = j + dir[k][1];if(nexti < 0||nexti >= grid.size()||nextj < 0||nextj >= grid[0].size()){continue;}if(visitedGrid.count(grid[nexti][nextj])) continue; // 添加过的岛屿不要重复添加count += gridNum[grid[nexti][nextj]];visitedGrid.insert(grid[nexti][nextj]);}}result = max(result,count);}}cout << result << endl;
}

思路：这道题是先计算各个岛屿的面积，然后遍历每一块海域，将这块添加的单元与其临近的岛屿面积进行相加求和，保存里面的最大值。

这道题真的是错死我了。。。这个涉及到很多细节，比如要考虑全是陆地的情况，要将每个岛屿进行面积和标号的映射，要在计算连接后的总面积的时候，将统计过的岛屿进行标记。。。我真的，太痛了。