LeetCode #1091. Shortest Path in Binary Matrix

题目描述：

In an N by N square grid, each cell is either empty (0) or blocked (1).

A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that:

• Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
• C_1 is at location (0, 0) (ie. has value grid[0][0])
• C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
• If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).

Return the length of the shortest such clear path from top-left to bottom-right.  If such a path does not exist, return -1.

Example 1:

Input: [[0,1],[1,0]]
Output: 2

Example 2:

Input: [[0,0,0],[1,1,0],[1,1,0]]
Output: 4

Note:

1. 1 <= grid.length == grid[0].length <= 100
2. grid[r][c] is 0 or 1

class Solution {
public:
    int shortestPathBinaryMatrix(vector<vector<int>>& grid) {
        if(grid[0][0]==1) return -1; // 需要单独判断起点为1的情况
        int n=grid.size();
        vector<vector<int>> dist(n,vector<int>(n,INT_MAX));
        dist[0][0]=1;
        vector<pair<int,int>> dirs={{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}};
        queue<pair<int,int>> q;
        q.push({0,0});
        while(!q.empty())
        {
            int i=q.front().first, j=q.front().second;
            q.pop();
            for(int k=0;k<8;k++)
            {
                int x=i+dirs[k].first, y=j+dirs[k].second;
                if(x<0||x>=n||y<0|y>=n||grid[x][y]==1) continue;
                if(dist[x][y]>dist[i][j]+1)
                {
                    dist[x][y]=dist[i][j]+1;
                    q.push({x,y});
                }
            }
        }
        if(dist[n-1][n-1]==INT_MAX) return -1;
        else return dist[n-1][n-1];
    }
};