zoukankan      html  css  js  c++  java

  • Tempter of the Bone

    Tempter of the Bone

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 31   Accepted Submission(s) : 3

    Font: Times New Roman | Verdana | Georgia

    Font Size: ← →

    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 
     1 #include <stdio.h>
 2 #include <stdlib.h>
 3 int Len_X,Len_Y,D,SIGN;
 4 int End_x,End_y;
 5 char Map[100][100];
 6 int DFS(int x,int y,int d,int D)
 7 {
 8     if(SIGN==1)return 1;
 9     if(x==End_x&&y==End_y&&d==D){SIGN+=1;return 1;}
10     if(Map[x][y]=='X')return 0;
11     else Map[x][y]='X';
12     if(d>=D) {return 0;}
13     if(x+1<Len_X&&Map[x+1][y]!='X')
14     {
15         DFS(x+1,y,d+1,D);
16         Map[x+1][y]='.';
17     }
18     if(x-1>=0&&Map[x-1][y]!='X')
19     {
20         DFS(x-1,y,d+1,D);
21         Map[x-1][y]='.';
22     }
23     if(y+1<Len_Y&&Map[x][y+1]!='X')
24     {
25         DFS(x,y+1,d+1,D);
26         Map[x][y+1]='.';
27     }
28     if(y-1>=0&&Map[x][y-1]!='X')
29     {
30         DFS(x,y-1,d+1,D);
31         Map[x][y-1]='.';
32     }
33     return 0;
34 }
35 
36 int main()
37 {
38     int Begin_x,Begin_y,i,j,flat,A,B;
39     while(scanf("%d%d%d",&Len_X,&Len_Y,&D)!=EOF)
40     {
41         if(Len_X==0&&Len_Y==0&&D==0)return 0;
42         getchar();
43         for(i=0;i<Len_X;i++)
44         {
45             for(j=0;j<Len_Y;j++)
46             {
47                 scanf("%c",&Map[i][j]);
48                 if(Map[i][j]=='S')
49                 {Begin_x=i;Begin_y=j;}
50                 if(Map[i][j]=='D')
51                 {End_x=i;End_y=j;}
52             }
53             getchar();
54         }
55         SIGN=0;
56         A=(End_y-Begin_y)>0?(End_y-Begin_y):-(End_y-Begin_y);
57         B=(End_x-Begin_x)>0?(End_x-Begin_x):-(End_x-Begin_x);
58         if(D%2==(A+B)%2)
59             DFS(Begin_x,Begin_y,0,D);
60         if(SIGN!=0)
61             printf("YES
");
62         else
63             printf("NO
");
64     }
65     return 0;
66 }
    View Code

     修改：2015.5.8

     1 #include <iostream>
 2 #include <stdio.h>
 3 #include <stdlib.h>
 4 using namespace std;
 5 int Len_X,Len_Y,D,SIGN;
 6 int End_x,End_y;
 7 char Map[100][100];
 8 void DFS(int x,int y,int d)
 9 {
10     //printf("%d %d %d
",x,y,D);getchar();
11     if(SIGN==1)return ;
12     if(x==End_x&&y==End_y&&d==D){SIGN=1;return ;}
13     if(d>=D) return ;
14     if(Map[x][y]=='X')return ;
15     else
16     {
17         Map[x][y]='X';
18         DFS(x+1,y,d+1);
19         DFS(x-1,y,d+1);
20         DFS(x,y+1,d+1);
21         DFS(x,y-1,d+1);
22         Map[x][y]='.';
23     }
24    
25     return ;
26 }
27 
28 int main()
29 {
30     int Begin_x,Begin_y,i,j,A,B;
31     while(scanf("%d%d%d",&Len_X,&Len_Y,&D)!=EOF)
32     {
33         if(Len_X==0&&Len_Y==0&&D==0)return 0;
34         Begin_x=-1;Begin_y=-1;End_x=-2;End_y=-2;
35         for(i=0;i<=Len_X+1;i++)
36         {
37             for(j=0;j<=Len_Y+1;j++)
38             {
39                 if(i==0||j==0||i==Len_X+1||j==Len_Y+1)Map[i][j]='X';
40                 else
41                 {
42                     scanf(" %c",&Map[i][j]);
43                     if(Map[i][j]=='S')
44                     {Begin_x=i;Begin_y=j;Map[i][j]='.';}
45                     if(Map[i][j]=='D')
46                     {End_x=i;End_y=j;Map[i][j]='.';}
47                 }
48 
49             }
50         }
51         SIGN=0;
52         A=(End_y-Begin_y)>0?(End_y-Begin_y):-(End_y-Begin_y);
53         B=(End_x-Begin_x)>0?(End_x-Begin_x):-(End_x-Begin_x);
54         if(D%2==(A+B)%2)
55         {
56             DFS(Begin_x,Begin_y,0);
57            /* DFS(Begin_x+1,Begin_y,1);
58             DFS(Begin_x-1,Begin_y,1);
59             DFS(Begin_x,Begin_y+1,1);
60             DFS(Begin_x,Begin_y-1,1);*/
61         }
62         if(SIGN)
63             printf("YES
");
64         else
65             printf("NO
");
66     }
67     return 0;
68 }
    View Code
     1 #include <iostream>
 2 #include <stdio.h>
 3 #include <stdlib.h>
 4 #define ABS(a,b)(((a)>=(b))?((a)-(b)):((b)-(a)))
 5 using namespace std;
 6 int Len_X,Len_Y,D,SIGN;
 7 int End_x,End_y;
 8 char Map[100][100];
 9 void DFS(int x,int y,int d)
10 {
11     //printf("%d %d %d
",x,y,D);getchar();
12     if(SIGN==1)return ;
13     if(x==End_x&&y==End_y&&d==D){SIGN=1;return ;}
14     if(d>=D) return ;
15     if(Map[x][y]=='X')return ;
16     else
17     {
18         Map[x][y]='X';
19         DFS(x+1,y,d+1);
20         DFS(x-1,y,d+1);
21         DFS(x,y+1,d+1);
22         DFS(x,y-1,d+1);
23         Map[x][y]='.';
24     }
25    
26     return ;
27 }
28 
29 int main()
30 {
31     int Begin_x,Begin_y,i,j,A,B;
32     while(scanf("%d%d%d",&Len_X,&Len_Y,&D)!=EOF)
33     {
34         if(Len_X==0&&Len_Y==0&&D==0)return 0;
35         Begin_x=-1;Begin_y=-1;End_x=-2;End_y=-2;
36         for(i=0;i<=Len_X+1;i++)
37         {
38             for(j=0;j<=Len_Y+1;j++)
39             {
40                 if(i==0||j==0||i==Len_X+1||j==Len_Y+1)Map[i][j]='X';
41                 else
42                 {
43                     scanf(" %c",&Map[i][j]);
44                     if(Map[i][j]=='S')
45                     {Begin_x=i;Begin_y=j;Map[i][j]='X';}/*标记已经走过*/
46                     if(Map[i][j]=='D')
47                     {End_x=i;End_y=j;}
48                 }
49 
50             }
51         }
52         SIGN=0;
53         A=ABS(End_x,Begin_x);
54         B=ABS(End_y,Begin_y);
55         if(D%2==(A+B)%2)
56         {
57             //DFS(Begin_x,Begin_y,0);
58             DFS(Begin_x+1,Begin_y,1);
59             DFS(Begin_x-1,Begin_y,1);
60             DFS(Begin_x,Begin_y+1,1);
61             DFS(Begin_x,Begin_y-1,1);
62         }
63         if(SIGN)
64             printf("YES
");
65         else
66             printf("NO
");
67     }
68     return 0;
69 }
    View Code


     

    转载请备注:
    **************************************
    * 作者: Wurq
    * 博客: https://www.cnblogs.com/Wurq/
    * Gitee: https://gitee.com/wurq
    **************************************
  • 相关阅读:
    SignalR客户端和服务端编写，winfrom端
    SQL 发送邮件msdb.dbo.sp_send_dbmail
    Stimulsoft打印内容横向，变成竖向，解决方法
    Chrome浏览器所有页面崩溃
    SQL存储过程直接传表
    c#SignalR一次发送最大数据量
    c#USB扫描枪，防止输入框手动输入
    SQLPrompt_9.5.5.9830破解版含注册机_永久激活不掉线
    SQL快捷键设置
    USB HDI 通信
  • 原文地址：https://www.cnblogs.com/Wurq/p/3750332.html
Copyright © 2011-2022 走看看