1293. Shortest Path in a Grid with Obstacles Elimination

问题：

给定n*m由0和1构成的二维数组。

0代表通路

1代表障碍

求从左上角格子(0,0)出发，要到达右下角格子(n-1,m-1)

可以最多遇到 k 个障碍，至少需要多少步能到达。

若不能做到，返回-1。

Example 1:
Input: 
grid = 
[[0,0,0],
 [1,1,0],
 [0,0,0],
 [0,1,1],
 [0,0,0]], 
k = 1
Output: 6
Explanation: 
The shortest path without eliminating any obstacle is 10. 
The shortest path with one obstacle elimination at position (3,2) is 6. Such path is (0,0) -> (0,1) -> (0,2) -> (1,2) -> (2,2) -> (3,2) -> (4,2).
 
Example 2:
Input: 
grid = 
[[0,1,1],
 [1,1,1],
 [1,0,0]], 
k = 1
Output: -1
Explanation: 
We need to eliminate at least two obstacles to find such a walk.
 

Constraints:
grid.length == m
grid[0].length == n
1 <= m, n <= 40
1 <= k <= m*n
grid[i][j] == 0 or 1
grid[0][0] == grid[m-1][n-1] == 0

　　

解法：BFS

• 状态：
  • 当前位置(x,y) 
  • 当前剩余可用遇障碍数 k
• 选择：
  • 上下左右4个方向，
    • 其中 k>0 时，grid[next.x][next.y] 可以==1。下一个状态 next.k--
    • 否则，grid[next.x][next.y] == 0 才可以选择。

代码参考：

class State {
public:
    int x;
    int y;
    int k;
    State(){}
    State(int x_1, int y_1, int k_1) {
        x=x_1;
        y=y_1;
        k=k_1;
    }
    string to_string() {
        return ::to_string(x) + "_" + ::to_string(y) + "_" + ::to_string(k);
    }
};
class Solution {
public:
    int m,n;
    int dir[5] = {1,0,-1,0,1};
    int shortestPath(vector<vector<int>>& grid, int k) {
        int step = 0;
        n = grid.size();
        m = grid[0].size();
        queue<State> q;
        unordered_set<string> visited;
        State st(0,0,k), cur, next;
        q.push(st);
        visited.insert(st.to_string());
        while(!q.empty()) {
            int sz = q.size();
            for(int k=0; k<sz; k++) {
                cur = q.front();
                q.pop();
                if(cur.x==n-1 && cur.y==m-1) return step;
                for(int d=1; d<5; d++) {
                    next.x = cur.x+dir[d-1];
                    next.y = cur.y+dir[d];
                    next.k = cur.k;
                    if(next.x<0 || next.y<0 || next.x>=n || next.y>=m) continue;
                    if(grid[next.x][next.y] == 1 && next.k<=0) continue;
                    else if(grid[next.x][next.y] == 1 && next.k>0) next.k--;
                    if(visited.insert(next.to_string()).second) {
                        q.push(next);
                    }
                }
            }
            step++;
        }
        return -1;
    }
};