这题有点像电子老鼠闯迷宫，也是用广搜来做的
我是用字符来输入的
const
dx:array[1..4]of longint=(1,-1,0,0);
dy:array[1..4]of longint=(0,0,1,-1);
var
px,py,x,y,n,s,tail:longint;
a:array[-1..2000,-1..2000]of char;
father:array[1..2000000]of longint;
state:array[1..1000000,1..2]of longint;
procedure init;
var
i,j:longint;
begin
readln(n);
for i:=1 to n do
begin
for j:=1 to n do
read(a[i,j]);
readln;
end;
readln(px,py,x,y);
end;
function check(x1,y1:longint):boolean;
begin
check:=true;
if a[x1,y1]='1' then check:=false;
if (x1>n)or(y1>n)or(y1<1)or(x1<1) then check:=false;
end;
procedure print(x:longint);
begin
if x=0 then exit;
inc(s);
print(father[x]);
end;
procedure bfs;
var
head,k:longint;
begin
head:=0;tail:=1;state[1,1]:=px;state[1,2]:=py;a[px,py]:='1';
father[1]:=0;
repeat
inc(head);
for k:=1 to 4 do
if check(state[head,1]+dx[k],state[head,2]+dy[k]) then
begin
inc(tail);
father[tail]:=head;
state[tail,1]:=state[head,1]+dx[k];
state[tail,2]:=state[head,2]+dy[k];
a[state[tail,1],state[tail,2]]:='1';
if (state[tail,1]=x)and(state[tail,2]=y) then
begin
print(father[tail]);
tail:=0;
writeln(s);
halt;
end;
end;
until head>=tail;
end;
begin
init;
bfs;
end.