<p>You are given an <code>n x n</code> binary matrix <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>&#39;s not connected to any other <code>1</code>&#39;s. There are <strong>exactly two islands</strong> in <code>grid</code>.</p>

<p>You may change <code>0</code>&#39;s to <code>1</code>&#39;s to connect the two islands to form <strong>one island</strong>.</p>

<p>Return <em>the smallest number of </em><code>0</code><em>&#39;s you must flip to connect the two islands</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,1,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There are exactly two islands in <code>grid</code>.</li>
</ul>

class Solution {
public:
vector<int> direction{-1, 0, 1, 0, -1};
// 主函数
int shortestBridge(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
queue<pair<int, int>> points;
// dfs寻找第一个岛屿， 并把1全部赋值为2
bool flipped = false;
for (int i = 0; i < m; ++i) {
if (flipped) break;
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
dfs(points, grid, m, n, i, j);
flipped = true;
break;
}
}
}
// bfs寻找第二个岛屿， 并把过程中经过的0赋值为2
int x, y;
int level = 0;
while (!points.empty()){
++level;
int n_points = points.size();
while (n_points--) {
auto [r, c] = points.front();
points.pop();
for (int k = 0; k < 4; ++k) {
x = r + direction[k], y = c + direction[k+1];
if (x >= 0 && y >= 0 && x < m && y < n) {
    if (grid[x][y] == 2) {
continue;
}
if (grid[x][y] == 1) {
return level;
}
points.push({x, y});
grid[x][y] = 2;
}
}
}
}
return 0;
}
// 辅函数
void dfs(queue<pair<int, int>>& points, vector<vector<int>>& grid, int m, int n
, int i, int j) {
if (i < 0 || j < 0 || i == m || j == n || grid[i][j] == 2) {
return;
}
if (grid[i][j] == 0) {
points.push({i, j});
return;
}
grid[i][j] = 2;
dfs(points, grid, m, n, i - 1, j);
dfs(points, grid, m, n, i + 1, j);
dfs(points, grid, m, n, i, j - 1);
dfs(points, grid, m, n, i, j + 1);
}
};
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