//Mark一下大神的解法
#include<iostream>
#include<vector>
using namespace std;
int N, M; //分别代表行和列
vector<vector<int>> maze;//迷宫矩阵
vector<vector<int>> path_temp;//存储当前路径,第一维表示位置
vector<vector<int>> path_best;//存储最佳路径
void MazeTrack(int i, int j)
{
maze[i][j] = 1;//表示当前节点已走,不可再走
path_temp.push_back({ i, j });//将当前节点加入到路径中
if (i == N - 1 && j == M - 1) //判断是否到达终点
if (path_best.empty() || path_temp.size() < path_best.size())
path_best = path_temp;
if (i - 1 >= 0 && maze[i - 1][j] == 0)//探索向上走是否可行
MazeTrack(i - 1, j);
if (i + 1 < N && maze[i + 1][j] == 0)//探索向下走是否可行
MazeTrack(i + 1, j);
if (j - 1 >= 0 && maze[i][j - 1] == 0)//探索向左走是否可行
MazeTrack(i, j - 1);
if (j + 1 < M && maze[i][j + 1] == 0)//探索向右走是否可行
MazeTrack(i, j + 1);
maze[i][j] = 0; //恢复现场,设为未走
path_temp.pop_back();
}
int main()
{
while (cin >> N >> M)
{
maze = vector<vector<int>>(N, vector<int>(M, 0));
path_temp.clear();
path_best.clear();
for (auto &i : maze)
for (auto &j : i)
cin >> j;
MazeTrack(0, 0);//回溯寻找迷宫最短通路
for (auto i : path_best)
cout << '(' << i[0] << ',' << i[1] << ')' << endl;//输出通路
}
return 0;
}
def FindPath(i,j,Matrix,N,M):
path=[[i,j]]
while(i!=N-1 or j!=M-1):#当前坐标不在下边界上,或者不在右边界上
if i<N-1 and Matrix[i+1][j]==0:#当前坐标的下侧坐标为0,即可以通行,则下移一位
i+=1
path.append([i,j])
elif j<M-1 and Matrix[i][j+1]==0:#当前坐标的右侧坐标为0,即可以通行,则右移一位
j+=1
path.append([i,j])
return path
while True:
try:
[N,M]=[int(i) for i in input().split()]
Matrix=[]
for i in range(N):
Matrix.append([int(i) for i in input().split()])
path=FindPath(0,0,Matrix,N,M)
for i in path:
print('(%d,%d)'%(i[0],i[1]))
except:
break