Description
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。
Input
Output
Sample Input
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
这题我们用的算法是广度优先搜素。
我们先定义head和tail两个指针,每一次判断这个点可不可以走,如果可以就将该定点记录下来。
代码如下:
const dx:array[1..4]of longint=(1,-1,0,0);
dy:array[1..4]of longint=(0,0,1,-1);
var a:array[1..100,1..100]of longint;
n,b1,b2,e1,e2,l,tail:longint;
state:array[1..100,1..2]of longint;
father:array[1..100]of longint;
procedure init;
var i,j:longint;
begin
readln(n);
readln(b1,b2,e1,e2);
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 a[x,y]=0 then check:=true
else check:=false;
end;
procedure print(x:longint);
begin
if x=0 then exit;
inc(l);
print(father[x]);
if x<>tail then write('(',state[x,1],',',state[x,2],')->')
else writeln('(',state[x,1],',',state[x,2],')');
end;
procedure bfs;
var head,i:longint;
begin
head:=0; tail:=1; state[1,1]:=b1; state[1,2]:=b2;
father[1]:=0;
repeat
inc(head);
for i:=1 to 4 do
if check(state[head,1]+dx[i],state[head,2]+dy[i])=true then
begin
inc(tail);
father[tail]:=head;
state[tail,1]:=state[head,1]+dx[i];
state[tail,2]:=state[head,2]+dy[i];
a[state[tail,1],state[tail,2]]:=1;
if (state[tail,1]=e1)and(state[tail,2]=e2) then
begin
print(tail);
tail:=0;
end;
end;
until head>=tail;
end;
begin
init;
bfs;
writeln(l);
end.