LEETCODE算法:542. 01 Matrix

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题目

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.

The distance between two adjacent cells is 1.

Example 1:

Input: [[0,0,0], [0,1,0], [0,0,0]]

Output: [[0,0,0], [0,1,0], [0,0,0]]

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

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

Note:

The number of elements of the given matrix will not exceed 10,000. There are at least one 0 in the given matrix. The cells are adjacent in only four directions: up, down, left and right.

题解

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//多元广度优先搜索，把所有为0的点加入到队列bfs即可，单源需要考虑的太多还容易超时。
class Solution {
    static constexpr int d[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int row=matrix.size();
        int line=matrix[0].size();
        vector<vector<int>> vis(row,vector<int>(line));
        vector<vector<int>> res(row,vector<int>(line));
        queue<pair<int,int>> q;
        for(int i=0;i<matrix.size();i++)
        {
            for(int j=0;j<matrix[i].size();j++)
            {
                if(matrix[i][j]==0){
                    q.push(make_pair(i,j));
                    vis[i][j]=1;
                }
            }
        }
        while(!q.empty())
        {
            auto tmp=q.front();
            q.pop();
            for(int i=0;i<4;i++)
            {
                int x=tmp.first+d[i][0];
                int y=tmp.second+d[i][1];
                if(x<row&&y<line&&x>=0&&y>=0&&!vis[x][y])
                {
                    q.push(make_pair(x,y));
                    res[x][y]=res[tmp.first][tmp.second]+1;
                    vis[x][y]=1;
                }
            }
        }
        return res;
    }

};