10pst求调必关

#include<iostream>
using namespace std;
const int N = 18;
int m, n, X[4] = {1, 0, 1, 0}, Y[4] = {0, -1, 0, 1}, x_0, y_0, x_m, y_n, ans[N * N][3];
bool mp[N][N], vis[N][N], win = 0;
void DFS(int x, int y, int cnt) {
	if (x < 1 || y < 1 || x > m || y > n || !mp[x][y] || vis[x][y])return;

	if (x == x_m && y == y_n) {
		for (int i = 1; i <= cnt; i++) {
			cout << '(' << ans[i][1] << ',' << ans[i][2] << ')';
			if (i != cnt)cout << "->";
		}
		cout << '\n';
		win = 1;
		return;
	}
	vis[x][y] = 1;
	ans[cnt][1] = x, ans[cnt][2] = y;
	for (int i = 0; i < 4; i++) {
		DFS(x + X[i], y + Y[i], cnt + 1);
	}
	vis[x][y] = 0;
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(nullptr), cout.tie(nullptr);
	cin >> m >> n;
	for (int i = 1; i <= m; i++)for (int j = 1; j <= n; j++)cin >> mp[i][j];
	cin >> x_0 >> y_0 >> x_m >> y_n;
	DFS(x_0, y_0, 1);
	if (!win)cout << "-1\n";
	return 0;
}