[ABC246E] Bishop

Problem Statement

We have an $N \times N$ chessboard. Let $(i, j)$ denote the square at the $i$-th row from the top and $j$-th column from the left of this board.

The board is described by $N$ strings $S_i$.

The $j$-th character of the string $S_i$, $S_{i,j}$, means the following.

• If $S_{i,j}=$ ., the square $(i, j)$ is empty.
• If $S_{i,j}=$ #, the square $(i, j)$ is occupied by a white pawn, which cannot be moved or removed.

We have put a white bishop on the square $(A_x, A_y)$.

Find the minimum number of moves needed to move this bishop from $(A_x, A_y)$ to $(B_x, B_y)$ according to the rules of chess (see Notes).

If it cannot be moved to $(B_x, B_y)$, report -1 instead.

Notes

A white bishop on the square $(i, j)$ can go to the following positions in one move.

• For each positive integer $d$, it can go to $(i+d,j+d)$ if all of the conditions are satisfied.

  • The square $(i+d,j+d)$ exists in the board.
  • For every positive integer $l \le d$, $(i+l,j+l)$ is not occupied by a white pawn.

• For each positive integer $d$, it can go to $(i+d,j-d)$ if all of the conditions are satisfied.

  • The square $(i+d,j-d)$ exists in the board.
  • For every positive integer $l \le d$, $(i+l,j-l)$ is not occupied by a white pawn.

• For each positive integer $d$, it can go to $(i-d,j+d)$ if all of the conditions are satisfied.

  • The square $(i-d,j+d)$ exists in the board.
  • For every positive integer $l \le d$, $(i-l,j+l)$ is not occupied by a white pawn.

• For each positive integer $d$, it can go to $(i-d,j-d)$ if all of the conditions are satisfied.

  • The square $(i-d,j-d)$ exists in the board.
  • For every positive integer $l \le d$, $(i-l,j-l)$ is not occupied by a white pawn.

Constraints

• $2 \le N \le 1500$
• $1 \le A_x,A_y \le N$
• $1 \le B_x,B_y \le N$
• $(A_x,A_y) \neq (B_x,B_y)$
• $S_i$ is a string of length $N$ consisting of . and #.
• $S_{A_x,A_y}=$ .
• $S_{B_x,B_y}=$ .

---

Input

Input is given from Standard Input in the following format:

$N$
$A_x$ $A_y$
$B_x$ $B_y$
$S_1$
$S_2$
$\vdots$
$S_N$

Output

Print the answer.

---

Sample Input 1

5
1 3
3 5
....#
...#.
.....
.#...
#....

Sample Output 1

3

We can move the bishop from $(1,3)$ to $(3,5)$ in three moves as follows, but not in two or fewer moves.

• $(1,3) \rightarrow (2,2) \rightarrow (4,4) \rightarrow (3,5)$

---

Sample Input 2

4
3 2
4 2
....
....
....
....

Sample Output 2

-1

There is no way to move the bishop from $(3,2)$ to $(4,2)$.

---

Sample Input 3

18
18 1
1 18
..................
.####.............
.#..#..####.......
.####..#..#..####.
.#..#..###...#....
.#..#..#..#..#....
.......####..#....
.............####.
..................
..................
.####.............
....#..#..#.......
.####..#..#..####.
.#.....####..#....
.####.....#..####.
..........#..#..#.
.............####.
..................

其实我们可以有另外一种理解方式：代价相当于在移动过程中转向的次数。我们可以记录上一步走的是那个方向，然后这一步的代价就是是否和上一步不等。

发现代价只有 0 或 1，01BFS 即可。

#include<bits/stdc++.h>
using namespace std;
const int N=1505,M=N*N*4,dx[]={1,-1,1,-1},dy[]={1,-1,-1,1};
int dp[N][N][4],n,sx,sy,ex,ey,l=1,r;
char s[N][N];
int mo(int x)
{
	return (x%M+M)%M;
}
int ok(int x,int y)
{
	return x>0&&y>0&&x<=n&&y<=n&&s[x][y]=='.';
}
struct node{
	int x,y,t;
}q[M];
int main()
{
	scanf("%d%d%d%d%d",&n,&sx,&sy,&ex,&ey);
	for(int i=1;i<=n;i++)
		scanf("%s",s[i]+1);
	memset(dp,0x7f,sizeof(dp));
	dp[sx][sy][0]=dp[sx][sy][2]=dp[sx][sy][3]=dp[sx][sy][1]=1;
	for(int i=0;i<4;i++)
		q[++r]=(node){sx,sy,i};
	while(l<=r)
	{
//		printf("%d %d\n",l,r);
		int x=q[mo(l)].x,y=q[mo(l)].y,t=q[mo(l)].t;
		++l;
		for(int i=0;i<4;i++)
		{
			int tx=x+dx[i],ty=y+dy[i];
			if(ok(tx,ty))
			{
				if(dp[x][y][t]+(i!=t)>=dp[tx][ty][i])
					continue;
				dp[tx][ty][i]=dp[x][y][t]+(i!=t);
				if(i==t)
				{
					--l;
					q[mo(l)]=(node){tx,ty,i};
				}
				else
				{
					++r;
					q[mo(r)]=(node){tx,ty,i};
				}
			}
		}
	}
//	for(int i=1;i<=n;i++)
//	{
//		for(int j=1;j<=n;j++)
//			printf("%d ",min(min(dp[i][j][0],dp[i][j][1]),min(dp[i][j][2],dp[i][j][3])));
//		putchar('\n');
//	}
	int ans=min(min(dp[ex][ey][0],dp[ex][ey][1]),min(dp[ex][ey][2],dp[ex][ey][3]));
	printf("%d",ans>2e9? -1:ans);
}