电子老鼠闯迷宫-ssl 1455
2016-12-22 16:29
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题意:
如下图12×12方格图,找出一条自入口(2,9)到出口(11,8)的最短路径。
Sample Output
分析:
这题数据范围大,用深搜会超时,要用广搜。
const
maxn=100;
warn=4;
ax:array[1..warn] of integer=(-1,0,1,0);
ay:array[1..warn] of integer=(0,1,0,-1);
var
a:array[1..maxn,1..maxn] of longint;
father:array[1..maxn*maxn] of longint;
state:array[1..maxn*maxn,1..2] of longint;
dx,dy,px,py,n,last,s:longint;
procedure init;
var i,j:longint;
begin
readln(n);
readln(dx,dy,px,py);
for i:=1 to n do
begin
for j:=1 to n do
read(a[i,j]);
readln;
end;
end;
function check(x,y:longint):boolean;
begin
if (x<1) or (x>12) or (y<1) or (y>12) then exit(false);
if a[x,y]=1 then exit(false);
exit(true);
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:longint;
begin
head:=0;tail:=1;state[1,1]:=dx;state[1,2]:=dy;
father[1]:=0;
repeat
inc(head);
for k:=1 to warn do
if check(state[head,1]+ax[k],state[head,2]+ay[k])
then begin
inc(tail);
father[tail]:=head;
state[tail,1]:=state[head,1]+ax[k];
state[tail,2]:=state[head,2]+ay[k];
a[state[tail,1],state[tail,2]]:=1;
if (state[tail,1]=px) and (state[tail,2]=py)
then begin
s:=0;
last:=tail;
print(tail);
writeln(s);
tail:=0;
end;
end;
until head>=tail;
end;
begin
init;
bfs;
end.
如下图12×12方格图,找出一条自入口(2,9)到出口(11,8)的最短路径。
12 //迷宫大小 2 9 11 8 //起点和终点 1 1 1 1 1 1 1 1 1 1 1 1 //邻接矩阵,0表示通,1表示不通 1 0 0 0 0 0 0 1 0 1 1 1 1 0 1 0 1 1 0 0 0 0 0 1 1 0 1 0 1 1 0 1 1 1 0 1 1 0 1 0 0 0 0 0 1 0 0 1 1 0 1 0 1 1 1 1 1 1 1 1 1 0 0 0 1 0 1 0 0 0 0 1 1 0 1 1 1 0 0 0 1 1 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 0 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Sample Output
(2,9)->(3,9)->(3,8)->(3,7)->(4,7)->(5,7)->(5,6)->(5,5)->(5,4)->(6,4)->(7,4)->(7,3)->(7,2)->(8,2)->(9,2)->(9,3)->(9,4)->(9,5)->(9,6)->(8,6)->(8,7)->(8,8)->(9,8)->(9,9)->(10,9)->(11,9)->(11,8) 27
分析:
这题数据范围大,用深搜会超时,要用广搜。
const
maxn=100;
warn=4;
ax:array[1..warn] of integer=(-1,0,1,0);
ay:array[1..warn] of integer=(0,1,0,-1);
var
a:array[1..maxn,1..maxn] of longint;
father:array[1..maxn*maxn] of longint;
state:array[1..maxn*maxn,1..2] of longint;
dx,dy,px,py,n,last,s:longint;
procedure init;
var i,j:longint;
begin
readln(n);
readln(dx,dy,px,py);
for i:=1 to n do
begin
for j:=1 to n do
read(a[i,j]);
readln;
end;
end;
function check(x,y:longint):boolean;
begin
if (x<1) or (x>12) or (y<1) or (y>12) then exit(false);
if a[x,y]=1 then exit(false);
exit(true);
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:longint;
begin
head:=0;tail:=1;state[1,1]:=dx;state[1,2]:=dy;
father[1]:=0;
repeat
inc(head);
for k:=1 to warn do
if check(state[head,1]+ax[k],state[head,2]+ay[k])
then begin
inc(tail);
father[tail]:=head;
state[tail,1]:=state[head,1]+ax[k];
state[tail,2]:=state[head,2]+ay[k];
a[state[tail,1],state[tail,2]]:=1;
if (state[tail,1]=px) and (state[tail,2]=py)
then begin
s:=0;
last:=tail;
print(tail);
writeln(s);
tail:=0;
end;
end;
until head>=tail;
end;
begin
init;
bfs;
end.
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