leetcode1284 转化为全零矩阵的最少反转次数

• m == mat.length
• n == mat[0].length
• 1 <= m <= 3
• 1 <= n <= 3
• mat[i][j] 是 0 或 1 。

BFS, 代码来自花花leetcode https://mp.weixin.qq.com/s/cxr4xp8iBRN6T2kgxWVTrw

可以再尝试DFS;

class Solution {
public:
    int minFlips(vector<vector<int>>& mat) {
        int m = mat.size();
        int n= mat[0].size();

        auto flip = [&](int &s, int x, int y){
            if(x<0 || x>=n || y<0 || y>=m) return;
            s^=(1<<y*n+x);
        };
        
        auto flipCell = [&](int s, int x, int y){
            int t=s;
            flip(t,x,y);
            flip(t,x-1,y);
            flip(t,x+1,y);
            flip(t,x,y+1);
            flip(t,x,y-1);
            return t;
        };

        int steps=0;
        queue<int> q;
        vector<int> seen(1<<(n*m));
        int s=0;
        for(int y=0;y<m;++y)
            for(int x=0;x<n;++x)
                s|=(mat[y][x]<<(y*n+x));
        if(s==0) return 0;
        q.push(s);
        seen[s]=1;

        while(!q.empty()){
            int size=q.size();
            while(size--){
                int ss=q.front();
                q.pop();
                for(int y=0;y<m;++y)
                    for(int x=0;x<n;++x){
                        int t=flipCell(ss,x,y);
                        if(seen[t]) continue;
                        if(t==0) return steps+1;
                        seen[t]=1;
                        q.push(t);
                    }
            }
            ++steps;
        }
        return -1;
    }
};