Leetcode 2556. Disconnect Path in a Binary Matrix by at Most One Flip
Woodyiiiiiii opened this issue · comments
这是一道关于割点的问题。我们可以用路径覆盖，第一次DFS将连接路径上的节点置为0，第二次继续判断是否能连通。
注意到路线的移动只能往右边和下边，所以我们不用考虑回环和可能覆盖其他连接路径的问题。
class Solution {
private final int[][] dirs = new int[][]{{0,1},{1,0}};
public boolean isPossibleToCutPath(int[][] grid) {
if (grid.length == 1 && grid[0].length == 2) return false;
boolean isConnected = dfs(grid, 0, 0);
grid[grid.length - 1][grid[0].length - 1] = 1;
return !isConnected || !dfs(grid, 0, 0);
}
private boolean dfs(int[][] grid, int i, int j) {
if (i == grid.length - 1 && j == grid[0].length - 1) {
return true;
}
grid[i][j] = 0;
for (int[] dir : dirs) {
int x = i + dir[0], y = j + dir[1];
if (x < 0 || x >= grid.length || y < 0 || y >= grid[0].length || grid[x][y] == 0) {
continue;
}
if (dfs(grid, x, y)) {
return true;
}
}
return false;
}
}