给定一个n * m 的矩阵，矩阵元素为0 或者 1，保证左上角和右下角元素均为1， 现在你可以从左上角出发沿上下左右四个方向向右下角前进，如果遇到0，就需要将 0 转换成为 1，请问最少需要转换多少个0才可以到达终点。

输入样例

4 5

1 0 0 0 0

1 0 1 1 1

1 0 1 0 1

1 1 1 0 1

注意：可能还有蛇皮路线，代码是自己写的可能有错误

#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
const int N = 1000;

int n, m, ans;
int g[N][N];
bool st[N][N];

void bfs(int x, int y)
{
    int dx[4] = {-1, 0, 1, 0};
    int dy[4] = {0, 1, 0, -1};

    queue<pair<int, int>> ones;
    queue<pair<int, int>> zeros;

    if(g[x][y] == 0) zeros.push({x, y});
    else ones.push({x, y});

    st[x][y] = true;

    while(zeros.size() || ones.size())
    {
        if(ones.size()) 
        {
            auto cur = ones.front();
            ones.pop();
            if(cur.first == n && cur.second == m) return;
            for(int i = 0; i < 4; i++)
            {
                int nx = cur.first + dx[i], ny = cur.second + dy[i];
                if(nx <= 0 || ny <= 0 || nx > n || ny > m) continue;
                if(st[nx][ny] == true) continue;
                st[nx][ny] = true;
                if(g[nx][ny] == 1) ones.push({nx, ny});
                else zeros.push({nx, ny});                
            }
        }
        else 
        {
            ans++;
            int s = zeros.size();
            for(int i = 0; i < s; i++)
            {
                auto cur = zeros.front();
                zeros.pop();
                for(int j = 0; j < 4; j++)
                {
                    int nx = cur.first + dx[j], ny = cur.second + dy[j];
                    if(nx <= 0 || ny <= 0 || nx > n || ny > m) continue;
                    if(st[nx][ny] == true) continue;
                    st[nx][ny] = true;
                    if(g[nx][ny] == 1) ones.push({nx, ny});
                    else zeros.push({nx, ny});                    
                }
            }            
        }
    }
}

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++)
            cin >> g[i][j];

    bfs(1, 1);

    cout << ans << endl;

    return 0;
}