dfs深搜，考虑了左上右下，也写了输出-1的代码

代码如下qwq：

#include <iostream>
using namespace std;
int n, m, sx, sy, ex, ey, nx, ny, te;
int xx[5] = {0, -1, 0, 1, 0};
int yy[5] = {0, 0, 1, 0, -1};
bool ac[16][16], vis[16][16], flag;
struct node{
	int x, y;
} a[400];
void print(int step){
	for(int i = 1; i < step; i++){
		cout << "(" << a[i].x << "," << a[i].y << ")" << "->";
	}
	cout << "(" << a[step].x << "," << a[step].y << ")" << endl;
}
void dfs(int p, int q, int step){
	ac[p][q] = false;
	a[step].x = p;
	a[step].y = q;
	if(p == ex && q == ey){
		flag = true;
		print(step);
	}
	for(int i = 1; i <= 4; i++){
		nx = p + xx[i];
		ny = q + yy[i];
		if(nx >= 1 && nx <= n && ny >= 1 && ny <= m && ac[nx][ny]){
			dfs(nx, ny, step + 1);
		}
	}
	ac[p][q] = true;
}
int main(){
	cin >> n >> m;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			cin >> te;
			if(te == 1){
				ac[i][j] = true;
			}
		}
	}
	cin >> sx >> sy >> ex >> ey;
	dfs(sx, sy, 1);
	if(!flag){
		cout << "-1" << endl;
	}
    return 0;
}