zoukankan      html  css  js  c++  java
  • 4.3.1 Tempter of the Bone

    Problem Description
    The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.

    The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.
     

    Input
    The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:

    \\\\\\\'X\\\\\\\': a block of wall, which the doggie cannot enter;
    \\\\\\\'S\\\\\\\': the start point of the doggie;
    \\\\\\\'D\\\\\\\': the Door; or
    \\\\\\\'.\\\\\\\': an empty block.

    The input is terminated with three 0\\\\\\\'s. This test case is not to be processed.
     

    Output

                For each test case, print in one line \\\\\\\"YES\\\\\\\" if the doggie can survive, or \\\\\\\"NO\\\\\\\" otherwise.
     

    Sample Input
    4 4 5
    S.X.
    ..X.
    ..XD
    ....
    3 4 5
    S.X.
    ..X.
    ...D
    0 0 0
     

    Sample Output
    NO
    YES
     

    Author
    ZHANG, Zheng
     

    Source
    ZJCPC2004
     

    Recommend
    JGShining

    思路:dfs,T了很多次,WA了很多次,开始没看见那个exactly T-ths - -||,很慢啊啊啊!

     1 #include <cstdio>
     2 #include <cstring>
     3 using namespace std;
     4 
     5 
     6 const int dx[]={0,0,1,-1};
     7 const int dy[]={1,-1,0,0};
     8 int n,m,t,xx,yy,ax,ay,tt=0,fc,cnt;
     9 char map[10][10];
    10 bool f[10][10],flag;
    11 
    12 void dfs(int x,int y)
    13 {
    14 int xx,yy;
    15 //printf("x:%d y:%d tt:%d \n",x,y,tt);
    16     for (int i=0;i<4;i++)
    17     {
    18         xx=dx[i]+x;
    19         yy=dy[i]+y;
    20       if (map[xx][yy]=='D') 
    21       {
    22     //        printf("tt:%d\n",tt);
    23             if (tt+1==t)
    24             {
    25                 flag=true;
    26                 return;
    27             }
    28     //      return;
    29       }
    30         if (xx>=0 && xx<n && yy>=0 && yy<m && not f[xx][yy] && map[xx][yy]=='.' && not flag && tt<=t) //&& tt<fc && tt<t && not flag)
    31         {
    32             f[xx][yy]=true;
    33             tt++;
    34             dfs(xx,yy);
    35             tt--;
    36         //    printf("---------------------\n");
    37             f[xx][yy]=false;
    38         }
    39     }
    40 }
    41 
    42 
    43 void init()
    44 {
    45     while (scanf("%d%d%d",&n,&m,&t)!=EOF)
    46     {
    47         cnt=0;
    48         if (n==0 && m==0 && t==0) break;
    49         for (int i=0;i<n;i++)
    50         {
    51             scanf("%s",map[i]);
    52             for (int j=0;j<m;j++)
    53             {
    54                 if (map[i][j]=='S')
    55                 {
    56                     ax=i;
    57                     ay=j;
    58                 }
    59                 if (map[i][j]=='.')
    60                     cnt++;
    61             }
    62         }
    63         memset(f,false,sizeof(f));
    64         flag=false;
    65         f[ax][ay]=true;
    66         tt=0;
    67         fc=0x3f3f3f;
    68         if (cnt+1>=t)
    69         dfs(ax,ay);
    70         if (flag==true)
    71             printf("YES\n");
    72         else printf("NO\n");
    73     }
    74 }
    75 
    76 int main ()
    77 {
    78 init();
    79 
    80 return 0;
    81 }
  • 相关阅读:
    OpenCASCADE Chamfer 3D Basics
    OpenCASCADE Chamfer 2D
    .NetCore 连接 Oracle 数据库,直接C# 或者 ORM框架(EFCore、XPO)
    心内科疾病指南
    HttpClient 调用 RestAPI 接口的用法
    在 Blazor 应用中使用 DevExtreme widgets
    2021 最近一次检查甘油三脂,验证苯扎贝特的效果。
    紫鹊界本味湘菜,
    如何Rest接口获取网上的股票数据,有哪些资源?-- 推荐Tushare金融数据
    优秀常用的「资源搜索网站」,收藏
  • 原文地址:https://www.cnblogs.com/cssystem/p/2836583.html
Copyright © 2011-2022 走看看