zoukankan      html  css  js  c++  java
  • SSL JudgeOnline 1455——电子老鼠闯迷宫

    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.
  • 相关阅读:
    把旧系统迁移到.Net Core 2.0 日记 (18) --JWT 认证(Json Web Token)
    把旧系统迁移到.Net Core 2.0 日记 (17) --多租户和SoftDelete
    swagger访问api, TypeError: Failed to fetch
    nop 4.1 Widget 探究- 视图组件
    Nop 4.1版本已经迁移到.net core2.1版本
    link标签和css引入方式
    script标签
    MIME 类型
    bae64编码
    chrome调试技巧和插件介绍
  • 原文地址:https://www.cnblogs.com/Comfortable/p/8412457.html
Copyright © 2011-2022 走看看