本题数据可能过水，我暴力bfs，$o(qn^4)$ 卡过去了，但是在本机上测试极限数据跑到 3 秒左右。
数据生成器：

#include<bits/stdc++.h>
using namespace std;
//mt19937_64 gen(time(0));
int vis[50][50];

int randlr(int l, int r){
	return rand() % (r - l + 1) + l;
}

int main(){
	srand(time(0)); 
	int n = 30, m = 30, q = 500;
	cout << n << ' ' << m << ' ' << q << endl;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			vis[i][j] = randlr(1, 10) > 1;
			cout << vis[i][j] << ' ';
		}
		cout << endl;
	}
	while(q--){
		int ex, ey, sx, sy, tx, ty;
		ex = randlr(1, n), ey = randlr(1, m);
		while(!vis[ex][ey]) ex = randlr(1, n), ey = randlr(1, m);
		sx = randlr(1, n), sy = randlr(1, m);
		while(!vis[sx][sy] || (sx == ex && sy == ey)) sx = randlr(1, n), sy = randlr(1, m);
		tx = randlr(1, n), ty = randlr(1, m);
		while(!vis[tx][ty]) tx = randlr(1, n), ty = randlr(1, m);
		cout << ex << ' ' << ey << ' ' << sx << ' ' << sy << ' ' << tx << ' ' << ty << endl;
	}
	return 0;
}

暴力代码：

int n, m, q;
int g[N][N];
int d[4][2] = {1, 0, 0, 1, -1, 0, 0, -1};
//int f[N][N][N][N]; // f(i,j,k,t) 从初始状态到 (空格在(i,j)，指定点在(k,t)) 的最小步数 

struct ti5 {int ex, ey, sx, sy, step;};

ti5 que[N * N * N * N];
int hd, tl;
bool vis[N][N][N][N];

void Solve(){
	cin >> n >> m >> q;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			cin >> g[i][j];
		}
	}
	int ex, ey, sx, sy, tx, ty;
	while(q--){
		memset(vis, 0, sizeof(vis));
		cin >> ex >> ey >> sx >> sy >> tx >> ty;
		//f[ex][ey][sx][sy] = 0;
		hd = 1, tl = 0;
		que[++tl] = (ti5){ex, ey, sx, sy, 0};
		vis[ex][ey][sx][sy] = 1;
		int ans = -1;
		while(hd <= tl){
			ti5 tmp = que[hd];
			hd++;
			if(tmp.sx == tx && tmp.sy == ty){
				ans = tmp.step;
				break;
			}
			for(register int i = 0; i < 4; i++){
				int xx = tmp.ex + d[i][0], yy = tmp.ey + d[i][1];
				if(xx < 1 || xx > n || yy < 1 || yy > m || g[xx][yy] == 0) continue;
				if(xx == tmp.sx && yy == tmp.sy){
					if(!vis[xx][yy][tmp.ex][tmp.ey]){
						que[++tl] = (ti5){xx, yy, tmp.ex, tmp.ey, tmp.step + 1};
						vis[xx][yy][tmp.ex][tmp.ey] = 1;
					} 
				}else{
					if(!vis[xx][yy][tmp.sx][tmp.sy]){
						que[++tl] = (ti5){xx, yy, tmp.sx, tmp.sy, tmp.step + 1};
						vis[xx][yy][tmp.sx][tmp.sy] = 1;
					}
				}
			}
		}
		cout << ans << endl;
	}
}