这题我是用广度优先搜索的方法来写的
head是头指针，tail是尾指针
我先找出下一个点，再判断这个点能不能走，能走就记录该点，到达终点时，就输出。
我是用递归的方法输出的
const
maxn=12;wayn=4;
dx:array[1..wayn]of longint=(-1,0,1,0);
dy:array[1..wayn]of longint=(0,1,0,-1);
var
px,py,qx,qy,s,last:longint;
a:array[0..maxn+1,0..maxn+1]of longint;
father:array[1..maxn*maxn]of longint;
state:array[1..maxn*maxn,1..2]of longint;
procedure init;
var
i,j,n:longint;
begin
readln(n);
readln(px,py,qx,qy);
for i:=1 to n do
begin
for j:=1 to n do
read(a[i,j]);
readln;
end;
end;
function check(x,y:integer):boolean;
begin
check:=true;
if (x<1)or(x>12)or(y<1)or(y>12) then check:=false;
if a[x,y]=1 then check:=false;
end;
procedure print(x:longint);
begin
if x=0 then exit;
inc(s);
print(father[x]);
if x<>last then write('(',state[x,1],',',state[x,2],')->') else
writeln('(',state[x,1],',',state[x,2],')');
end;
procedure bfs;
var
tail,head,k,i:integer;
begin
head:=0;tail:=1;state[1,1]:=px;state[1,2]:=py;
father[1]:=0;
repeat
inc(head);
for k:=1 to wayn 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]=qx)and(state[tail,2]=qy) then
begin
s:=0;
last:=tail;
print(tail);
writeln(s);
tail:=0;
end;
end;
until head>=tail;
end;
begin
init;
bfs;
end.
这题是我第一次用广度优先搜索的方法来写程序，希望我能把广度优先搜索学会。