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934. Shortest Bridge

bfs O(mn) time
先找到一个岛变成3同时找到岛的边缘海变成2
沿着边缘海扩张直到找到1为止
两个bfs即可

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class Solution {
public:
int shortestBridge(vector<vector<int>>& A) {
m = A.size(), n = A[0].size();
auto [r, c] = getRC(A);
int res = 0;
auto &&q = update(A, r, c);
while (!q.empty()) {
++res;
for (int i = q.size(); i > 0; --i) {
auto [r, c] = q.front(); q.pop();
for (int j = 0; j < 4; ++j) {
int rr = r + dr[j], cc = c + dc[j];
if (!isValid(rr, cc)) continue;
if (A[rr][cc] == 0) {
q.emplace(rr, cc);
A[rr][cc] = 2;
} else if (A[rr][cc] == 1) return res;
}
}
}
return res;
}

pair<int, int> getRC(vector<vector<int>> &A) {
for (int r = 0; r < m; ++r) {
for (int c = 0; c < n; ++c) {
if (A[r][c] == 1) {
return {r, c};
}
}
}
return {0, 0};
}

queue<pair<int, int>> update(vector<vector<int>> &A, int r, int c) {
queue<pair<int, int>> q{{{r, c}}}, res;
A[r][c] = 3;
while (!q.empty()) {
auto [r, c] = q.front(); q.pop();
for (int i = 0; i < 4; ++i) {
int rr = r + dr[i], cc = c + dc[i];
if (isValid(rr, cc)) {
if (A[rr][cc] == 1) {
q.emplace(rr, cc);
A[rr][cc] = 3;
} else if (A[rr][cc] == 0) {
res.emplace(rr, cc);
A[rr][cc] = 2;
}
}
}
}
return res;
}

bool isValid(int r, int c) {
return 0 <= r && r < m && 0 <= c && c < n;
}

int m, n, dr[4] = {0, 0, 1, -1}, dc[4] = {1, -1, 0, 0};
};