*算法训练（leetcode）第四十五天 | 101. 孤岛的总面积、102. 沉没孤岛、103. 水流问题、104. 建造最大岛屿

101. 孤岛的总面积

题目地址

本题要求不与矩阵边缘相连的孤岛的总面积。先将与四个边缘相连的岛屿变为海洋，再统计剩余的孤岛的总面积。无需再标识访问过的结点，因为访问过后都变为海洋了。

时间复杂度： $O(n^2)$
空间复杂度： $O(n^2)$

DFS

// c++
#include<bits/stdc++.h>
using namespace std;
int direction[4][2] = {0, 1, 0, -1, -1, 0, 1, 0};
int result = 0;void pre_dfs(vector<vector<int>> &matrix, int x, int  y){matrix[x][y] = 0;result++;for(int i=0; i<4; i++){int nextx = x + direction[i][0];int nexty = y + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix[0].size() || nextx<0 || nexty<0) continue;if(matrix[nextx][nexty]){matrix[nextx][nexty] = 0;pre_dfs(matrix, nextx, nexty);}}
}int main(){int n,m;cin>>n>>m;vector<vector<int>> matrix(n, vector<int>(m, 0));for(int i=0; i<n; i++){for(int j=0; j<m; j++){cin>>matrix[i][j];}}for(int i=0; i<n; i++) {if(matrix[i][0]){pre_dfs(matrix, i, 0);}if(matrix[i][m-1]){pre_dfs(matrix, i, m-1);}}for(int j=0; j<m; j++){if(matrix[0][j]){pre_dfs(matrix, 0, j);}if(matrix[n-1][j]){pre_dfs(matrix, n-1, j);   }}result = 0;for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(matrix[i][j]){pre_dfs(matrix, i, j);   }}}cout<<result;return 0;
}

BFS

// c++
#include<bits/stdc++.h>
using namespace std;
int direction[4][2] = {0, 1, 0, -1, -1, 0, 1, 0};
int result = 0;void pre_dfs(vector<vector<int>> &matrix, int x, int  y){matrix[x][y] = 0;result++;for(int i=0; i<4; i++){int nextx = x + direction[i][0];int nexty = y + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix[0].size() || nextx<0 || nexty<0) continue;if(matrix[nextx][nexty]){matrix[nextx][nexty] = 0;pre_dfs(matrix, nextx, nexty);}}
}void pre_bfs(vector<vector<int>> &matrix, int x, int  y){queue<pair<int, int>> que;que.push({x, y});matrix[x][y] = 0;result++;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 + direction[i][0];int nexty = cury + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix[0].size() || nextx<0 || nexty<0) continue;if(matrix[nextx][nexty]){matrix[nextx][nexty] = 0;result++;que.push({nextx, nexty});}}}
}int main(){int n,m;cin>>n>>m;vector<vector<int>> matrix(n, vector<int>(m, 0));for(int i=0; i<n; i++){for(int j=0; j<m; j++){cin>>matrix[i][j];}}/*// dfsfor(int i=0; i<n; i++) {if(matrix[i][0]){pre_dfs(matrix, i, 0);}if(matrix[i][m-1]){pre_dfs(matrix, i, m-1);}}for(int j=0; j<m; j++){if(matrix[0][j]){pre_dfs(matrix, 0, j);}if(matrix[n-1][j]){pre_dfs(matrix, n-1, j);   }}*/// bfsfor(int i=0; i<n; i++) {if(matrix[i][0]){pre_bfs(matrix, i, 0);}if(matrix[i][m-1]){pre_bfs(matrix, i, m-1);}}for(int j=0; j<m; j++){if(matrix[0][j]){pre_bfs(matrix, 0, j);}if(matrix[n-1][j]){pre_bfs(matrix, n-1, j);   }}result = 0;for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(matrix[i][j]){// pre_dfs(matrix, i, j);   pre_bfs(matrix, i, j);}}}cout<<result;return 0;
}

102. 沉没孤岛

题目地址

本题是上一题的反向操作

先把非孤岛做访问标记，再对剩余陆地进行操作。

时间复杂度： $O(n^2)$
空间复杂度： $O(n^2)$

DFS

// c++
#include<bits/stdc++.h>
using namespace std;
int direction[][2] = {0, 1, 0, -1, -1, 0, 1, 0};
void pre_dfs(const vector<vector<int>>& matrix,vector<vector<bool>>& visited,int x, int y){visited[x][y] = true;for(int i=0; i<4; i++){int nextx = x + direction[i][0];int nexty = y + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix.size() ||nextx<0 || nexty<0) continue;if(matrix[nextx][nexty] && !visited[nextx][nexty]){// visited[nextx][nexty] = true;pre_dfs(matrix, visited, nextx, nexty);}}
}void dfs(vector<vector<int>>& matrix,vector<vector<bool>>& visited,int x, int y){matrix[x][y] = 0;for(int i=0; i<4; i++){int nextx = x + direction[i][0];int nexty = y + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix.size() ||nextx<0 || nexty<0) continue;if(matrix[nextx][nexty] && !visited[nextx][nexty]){visited[nextx][nexty] = true;dfs(matrix, visited, nextx, nexty);}}
}
int main(){int n,m;cin>>n>>m;vector<vector<int>> matrix(n, vector<int>(m, 0));vector<vector<bool>> visited(n, vector<bool>(m, false));for(int i=0; i<n; i++){for(int j=0; j<m; j++){cin>>matrix[i][j];}}for(int i=0; i<n; i++){if(matrix[i][0] && !visited[i][0]) pre_dfs(matrix, visited, i, 0);if(matrix[i][m-1] && !visited[i][m-1]) pre_dfs(matrix, visited, i, m-1);}for(int j=0; j<m; j++){if(matrix[0][j] && !visited[0][j]) pre_dfs(matrix, visited, 0, j);if(matrix[n-1][j] && !visited[n-1][j]) pre_dfs(matrix, visited, n-1, j);}for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(matrix[i][j] && !visited[i][j]){visited[i][j] = true;dfs(matrix,visited, i, j);}}for(int j=0; j<m; j++) cout<<matrix[i][j]<<" ";cout<<endl;}return 0;
}

BFS

//c++
#include<bits/stdc++.h>
using namespace std;
int direction[][2] = {0, 1, 0, -1, -1, 0, 1, 0};
void pre_dfs(const vector<vector<int>>& matrix,vector<vector<bool>>& visited,int x, int y){visited[x][y] = true;for(int i=0; i<4; i++){int nextx = x + direction[i][0];int nexty = y + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix.size() ||nextx<0 || nexty<0) continue;if(matrix[nextx][nexty] && !visited[nextx][nexty]){// visited[nextx][nexty] = true;pre_dfs(matrix, visited, nextx, nexty);}}
}void dfs(vector<vector<int>>& matrix,vector<vector<bool>>& visited,int x, int y){matrix[x][y] = 0;for(int i=0; i<4; i++){int nextx = x + direction[i][0];int nexty = y + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix.size() ||nextx<0 || nexty<0) continue;if(matrix[nextx][nexty] && !visited[nextx][nexty]){visited[nextx][nexty] = true;dfs(matrix, visited, nextx, nexty);}}
}void pre_bfs(const vector<vector<int>>& matrix,vector<vector<bool>>& visited,int x, int y){visited[x][y] = true;queue<pair<int, int>> que;que.push({x,y});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 + direction[i][0];int nexty = cury + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix.size() ||nextx<0 || nexty<0) continue;if(matrix[nextx][nexty] && !visited[nextx][nexty]){visited[nextx][nexty] = true;que.push({nextx, nexty});}}}
}void bfs(vector<vector<int>>& matrix,vector<vector<bool>>& visited,int x, int y){visited[x][y] = true;matrix[x][y] = 0;queue<pair<int, int>> que;que.push({x,y});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 + direction[i][0];int nexty = cury + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix.size() ||nextx<0 || nexty<0) continue;if(matrix[nextx][nexty] && !visited[nextx][nexty]){visited[nextx][nexty] = true;matrix[nextx][nexty] = 0;que.push({nextx, nexty});}}}
}
int main(){int n,m;cin>>n>>m;vector<vector<int>> matrix(n, vector<int>(m, 0));vector<vector<bool>> visited(n, vector<bool>(m, false));for(int i=0; i<n; i++){for(int j=0; j<m; j++){cin>>matrix[i][j];}}/*// dfsfor(int i=0; i<n; i++){if(matrix[i][0] && !visited[i][0]) pre_dfs(matrix, visited, i, 0);if(matrix[i][m-1] && !visited[i][m-1]) pre_dfs(matrix, visited, i, m-1);}for(int j=0; j<m; j++){if(matrix[0][j] && !visited[0][j]) pre_dfs(matrix, visited, 0, j);if(matrix[n-1][j] && !visited[n-1][j]) pre_dfs(matrix, visited, n-1, j);}*/// bfsfor(int i=0; i<n; i++){if(matrix[i][0] && !visited[i][0]) pre_bfs(matrix, visited, i, 0);if(matrix[i][m-1] && !visited[i][m-1]) pre_bfs(matrix, visited, i, m-1);}for(int j=0; j<m; j++){if(matrix[0][j] && !visited[0][j]) pre_bfs(matrix, visited, 0, j);if(matrix[n-1][j] && !visited[n-1][j]) pre_bfs(matrix, visited, n-1, j);}for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(matrix[i][j] && !visited[i][j]){// visited[i][j] = true;// dfs(matrix,visited, i, j);bfs(matrix,visited, i, j);}}for(int j=0; j<m; j++) cout<<matrix[i][j]<<" ";cout<<endl;}return 0;
}

*103. 水流问题

题目地址

使用两个标识访问的数组分别从两组边界出发进行dfs遍历，使用从低向高流(反向流)来分别记录两组边界的结点。最后两组边界的交集就是本题答案。
思路

时间复杂度： $O(m\ast n)$
空间复杂度： $O(m\ast n)$

// c++
#include<bits/stdc++.h>
using namespace std;
int direction[][2] = {0, 1, 0, -1, -1, 0, 1, 0};
void dfs(const vector<vector<int>> &matrix, vector<vector<bool>> &visited,int x, int y){visited[x][y] = true;for(int i=0; i<4; i++){int nextx = x + direction[i][0];int nexty = y + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix[0].size() || nextx<0 || nexty<0) continue;if(matrix[x][y]>matrix[nextx][nexty]) continue;if(!visited[nextx][nexty]) dfs(matrix, visited, nextx, nexty);}
}int main(){int n,m;cin>>n>>m;vector<vector<int>> matrix(n, vector<int>(m, 0));vector<vector<bool>> first(n, vector<bool>(m, false));vector<vector<bool>> second(n, vector<bool>(m, false));for(int i=0; i<n; i++){for(int j=0; j<m; j++){cin>>matrix[i][j];}}for(int i=0; i<n; i++){dfs(matrix, first, i, 0);dfs(matrix, second, i, m-1);}for(int j=0; j<m; j++){dfs(matrix, first, 0, j);dfs(matrix, second, n-1, j);}for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(first[i][j] && second[i][j]) cout<<i<<" "<<j<<endl;}}return 0;   
}

*104. 建造最大岛屿

题目地址

分两阶段。

1. 先对地图中的陆地进行一次遍历，对每一个岛屿进行标记，同时计算每个岛屿的面积。
2. 再对地图中的水域进行一次遍历，计算每一个水域变成陆地后的连续陆地面积，就是把水域四个方向的陆地面积求和，再加上1表示当前水域变为陆地。

题解

时间复杂度： $O(n^2)$
空间复杂度： $O(n^2)$

// c++
#include<bits/stdc++.h>
using namespace std;
int direction[4][2] = {0, 1, 0, -1, -1, 0, 1, 0};void dfs(vector<vector<int>> &matrix, vector<vector<bool>> &visited, int x, int  y, int &count, int mark){visited[x][y] = true;matrix[x][y] = mark;count++;for(int i=0; i<4; i++){int nextx = x + direction[i][0];int nexty = y + direction[i][1];if(nextx>=matrix.size() || nexty>=matrix[0].size() || nextx<0 || nexty<0) continue;if(matrix[nextx][nexty] == 1 && !visited[nextx][nexty]){dfs(matrix, visited, nextx, nexty, count, mark);}}
}int main(){int n,m;cin>>n>>m;vector<vector<int>> matrix(n, vector<int>(m, 0));for(int i=0; i<n; i++){for(int j=0; j<m; j++){cin>>matrix[i][j];}}vector<vector<bool>> visited(n, vector<bool>(m, false));// 记录岛屿面积unordered_map<int, int> markCount;int mark = 2;int count;bool isAllGrid = true;// 第一阶段：记录岛屿大小for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(matrix[i][j] == 0) isAllGrid = false;if(matrix[i][j] == 1 && !visited[i][j]){count = 0;dfs(matrix, visited, i, j, count, mark);markCount[mark] = count;mark++;}}}if(isAllGrid) {cout<<n*m<<endl;return 0;}int result = 0;// 第二阶段for(int i=0; i<n; i++){for(int j=0; j<m; j++){// 非陆地 计算把当前水域变为陆地的最大面积if(matrix[i][j] == 0){unordered_set<int> lands;// 把当前水域变为陆地count = 1;// 遍历当前水域的四个方向 把四个方向的陆地面积都加起来for(int k=0; k<4; k++){int nextx = i + direction[k][0];int nexty = j + direction[k][1];if(nextx>=matrix.size() || nexty>=matrix[0].size() || nextx<0 || nexty<0) continue;// 当前陆地已加入if(lands.count(matrix[nextx][nexty])>0) continue;count += markCount[matrix[nextx][nexty]];// 标记已加入的节点lands.insert(matrix[nextx][nexty]);}}result = max(result, count);}}cout<<result;return 0;
}