# 图论第2天----第1020题、第130题

​ 又继续开始修行，把图论这块补上，估计要个5-6天时间。

一、第1020题–飞地的数量

​ 跟前面做的思路差不多，其实有另外一种思路。我这里是在遍历每块岛屿时，判断是否四周都是水，进而再统计数量。另一种思路，与130思路差不多，从边上开始遍历，要遍历2遍。

class Solution {
public:int count;int result = 0;bool flag;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){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()){flag = true;continue;}if(!visited[nextx][nexty] && grid[nextx][nexty] == 1){visited[nextx][nexty] = true;count++;dfs(grid, visited, nextx, nexty);}}}int numEnclaves(vector<vector<int>>& grid) {int n = grid.size();int m = grid[0].size();vector<vector<bool>> visited(n, vector<bool>(m, false));for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(!visited[i][j] && grid[i][j] == 1){visited[i][j] = true;flag = false;count = 1;dfs(grid, visited, i, j);cout << flag << ' ' << count << endl;if (!flag) result += count;}}}return result;}
};

二、第130题–被围绕的区域

​ 遍历2遍。第一遍从边上开始遍历，把边上的岛屿做标记，记为’A’。这样边上的岛屿（‘A’）与中间的岛屿(‘O’)区分开了。第二遍遍历时，每个元素都遍历到，把’O’变为’X’，把’A’变为’O’，就把边上边上的岛屿留下了。

class Solution {
public:int dir[4][2] = {0,1,1,0,0,-1,-1,0};void dfs(vector<vector<char>>& board, int x, int y){board[x][y] = 'A';for(int i=0; i<4; i++){int nextx = x + dir[i][0];int nexty = y + dir[i][1];if(nextx <0 || nexty <0 || nextx >=board.size() || nexty >=board[0].size()) continue;if(board[nextx][nexty] == 'X' || board[nextx][nexty] == 'A') continue;dfs(board, nextx, nexty);}}void solve(vector<vector<char>>& board) {int n = board.size();int m = board[0].size();for(int i=0; i<n; i++){if(board[i][0] == 'O') dfs(board, i, 0);if(board[i][m-1] == 'O') dfs(board, i, m-1);} for(int j=0; j<m; j++){if(board[0][j] == 'O') dfs(board, 0, j);if(board[n-1][j] == 'O') dfs(board, n-1, j);}for(int i=0; i<n; i++){for(int j=0; j<m; j++){if(board[i][j] == 'O') board[i][j] = 'X';if(board[i][j] == 'A') board[i][j] = 'O';}}}
};