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

    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
    代码:
    数据都对,但是没通过..就这思路吧
     1 #include<cstdio>
     2 #include<cstring>
     3 #include<queue>
     4 #include<algorithm>
     5 using namespace std;
     6 struct xy
     7 {int x,y,step;
     8     
     9 }pr,ne;
    10 int n,m,t,stx,sty,enddx,enddy;
    11 char a[7][7];
    12 int v[7][7];
    13 int fx[4]={0,0,1,-1},fy[4]={1,-1,0,0};
    14 int bfs()
    15 {pr.x=stx;
    16  pr.y=sty;
    17  pr.step=0;
    18  memset(v,0,sizeof(v));
    19  v[stx][sty]=1;
    20  queue<xy>q;
    21  q.push(pr);
    22  while(!q.empty())
    23     {pr=q.front();
    24      q.pop();
    25      for(int i=0;i<4;i++)
    26        {ne.x=pr.x+fx[i];
    27         ne.y=pr.y+fy[i];
    28         ne.step=pr.step+1;
    29         if(!(ne.x>=0&&ne.y>=0&&ne.x<n&&ne.y<m&&!v[ne.x][ne.y]&&a[ne.x][ne.y]!='X'))
    30              continue;
    31             if(ne.x==enddx&&ne.y==enddy)
    32             {return ne.step;
    33                 
    34              }
    35          q.push(ne);      
    36          v[ne.x][ne.y]=1;
    37               
    38        }    
    39     }
    40  return -1;    
    41 }
    42 int main()
    43 {
    44     while(~scanf("%d %d %d",&n,&m,&t)&&(n||m||t))
    45     {for(int i=0;i<n;i++)
    46        scanf("%s",a[i]);
    47      for(int i=0;i<n;i++)
    48        for(int j=0;j<m;j++)
    49          {if(a[i][j]=='S')
    50             {stx=i;
    51              sty=j;
    52                 
    53             }
    54           if(a[i][j]=='D')
    55             {enddx=i;
    56              enddy=j;
    57                 
    58             }
    59                   
    60              
    61          } 
    62      if(bfs()<=t&&bfs()>=0) printf("YES
    ");
    63      else printf("NO
    ");      
    64         
    65     }
    66     return 0;
    67 }
  • 相关阅读:
    angular2 UT 导入 jquery问题解决
    css超过指定宽度用...表示
    karma-coverage通过浏览器显示
    angular2复选框及其按钮
    前端分页控制
    input复选框checkbox默认样式纯css修改
    弧形侧边栏
    浅谈软件测试
    随笔1
    java注解小记
  • 原文地址:https://www.cnblogs.com/hss-521/p/7270346.html
Copyright © 2011-2022 走看看