// 2577. Minimum Time to Visit a Cell In a Grid #include "leetcode.h" /* * you are given an 'm x n' matrix 'grid' consisting of non-negative integers * where 'grid[row][col]' represents the minimum time required to be able to * visit the cell '(row, col)', which means you can visit the cell '(row, col)' * only when the time you visit it is greater than or equal to the value. you * are standing in the top left cell of the matrix in the 0th second and you * must move to any adjacent cell in the four directions: up, down, left, and * right. each move you make takes 1 second. return the minimum time required in * which you can visit the bottom right cell of the matrix. if you cannot visit * the bottomright cell, then return -1. */ class Solution { public: int minimumTime(vector> &grid) { if (grid[0][1] > 1 && grid[1][0] > 1) return -1; int m = grid.size(), n = grid[0].size(); vector> dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; vector> vis(m, vector(n, false)); priority_queue, vector>, greater<>> pq; pq.push({grid[0][0], 0, 0}); while (!pq.empty()) { int time = pq.top()[0], row = pq.top()[1], col = pq.top()[2]; pq.pop(); if (row == m - 1 && col == n - 1) return time; if (vis[row][col]) continue; vis[row][col] = true; for (auto i : dir) { int r = row + i[0], c = col + i[1]; if (r < 0 || r >= m || c < 0 || c >= n || vis[r][c]) continue; int wait = !((grid[r][c] - time) % 2); pq.push({max(grid[r][c] + wait, time + 1), r, c}); } } return -1; } }; int main() { Solution obj; vvd g1 = {{0, 1, 3, 2}, {5, 1, 2, 5}, {4, 3, 8, 6}}, g2 = {{0, 2, 4}, {3, 2, 1}, {1, 0, 4}}; printf("%d\n", obj.minimumTime(g1)); // expect: 7 printf("%d\n", obj.minimumTime(g2)); // expect: -1 }