SSL JudgeOnline 1455——电子老鼠闯迷宫

来源：互联网发布：网络舆论的传播分析编辑：程序博客网时间：2024/06/05 09:22

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。

这里写图片描述
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.

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