zoukankan      html  css  js  c++  java
  • HDU_1240——三围空间BFS

    好吧,BFS用模版还是比较好理解的。这代码能AC- -

    Problem Description
    You're in space.
    You want to get home.
    There are asteroids.
    You don't want to hit them.
     
    Input
    Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.

    A single data set has 5 components:

    Start line - A single line, "START N", where 1 <= N <= 10.

    Slice list - A series of N slices. Each slice is an N x N matrix representing a horizontal slice through the asteroid field. Each position in the matrix will be one of two values:

    'O' - (the letter "oh") Empty space

    'X' - (upper-case) Asteroid present

    Starting Position - A single line, "A B C", denoting the <A,B,C> coordinates of your craft's starting position. The coordinate values will be integers separated by individual spaces.

    Target Position - A single line, "D E F", denoting the <D,E,F> coordinates of your target's position. The coordinate values will be integers separated by individual spaces.

    End line - A single line, "END"

    The origin of the coordinate system is <0,0,0>. Therefore, each component of each coordinate vector will be an integer between 0 and N-1, inclusive.

    The first coordinate in a set indicates the column. Left column = 0.

    The second coordinate in a set indicates the row. Top row = 0.

    The third coordinate in a set indicates the slice. First slice = 0.

    Both the Starting Position and the Target Position will be in empty space.

     
    Output
    For each data set, there will be exactly one output set, and there will be no blank lines separating output sets.

    A single output set consists of a single line. If a route exists, the line will be in the format "X Y", where X is the same as N from the corresponding input data set and Y is the least number of moves necessary to get your ship from the starting position to the target position. If there is no route from the starting position to the target position, the line will be "NO ROUTE" instead.

    A move can only be in one of the six basic directions: up, down, left, right, forward, back. Phrased more precisely, a move will either increment or decrement a single component of your current position vector by 1.

     
    Sample Input
    START 1 O 0 0 0 0 0 0 END START 3 XXX XXX XXX OOO OOO OOO XXX XXX XXX 0 0 1 2 2 1 END START 5 OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO XXXXX XXXXX XXXXX XXXXX XXXXX OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO OOOOO 0 0 0 4 4 4 END
     
    Sample Output
    1 0 3 4 NO ROUTE
     1 #include <cstdio>
     2 #include <cstring>
     3 #include <queue>
     4 #define MAX 11
     5 using namespace std;
     6 
     7 typedef struct node
     8 {
     9    int x,y,z;
    10    int move;
    11 }point;
    12 point start,end;
    13 
    14 const int dir[7][3]={{0,0,0},{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
    15 int N;
    16 char str[MAX],map[MAX][MAX][MAX];
    17 
    18 void Bfs(void);
    19 int main()
    20 {
    21    int i,j,k;
    22    while(scanf("%s",str)!=EOF)
    23       {
    24          scanf("%d",&N);
    25          memset(map,0,sizeof(map));
    26          for(i=0;i<N;i++)
    27             for(j=0;j<N;j++)
    28                scanf("%s",map[i][j]);
    29          start.move=0;
    30          scanf("%d %d %d",&start.x,&start.y,&start.z);
    31          scanf("%d %d %d",&end.x,&end.y,&end.z);
    32          scanf("%s",str);
    33          Bfs();
    34       }
    35    return 0;
    36 }
    37 
    38 void Bfs()
    39 {
    40    queue<point>p,p_free;
    41    p.push(start);
    42    point temp,next;
    43    while(!p.empty())
    44       {
    45          temp=p.front();
    46          p.pop();
    47          for(int i=0;i<7;i++)
    48             {
    49                next.x=temp.x+dir[i][0];
    50                next.y=temp.y+dir[i][1];
    51                next.z=temp.z+dir[i][2];
    52                if(i==0)
    53                   next.move=temp.move;
    54                else
    55                   next.move=temp.move+1;
    56                if(next.x>=0&&next.x<N && next.y>=0&&next.y<N && next.z>=0&&next.z<N)
    57                   if(map[next.z][next.y][next.x]=='O')
    58                      {
    59                         map[next.z][next.y][next.x]='@';
    60                         p.push(next);
    61                         if(next.x==end.x && next.y==end.y && next.z==end.z)
    62                            {
    63                               printf("%d %d\n",N,next.move);
    64                               return;
    65                            }
    66                      }
    67             }   
    68       }
    69    printf("NO ROUTE\n");
    70    return;
    71 }
    72 /*
    73 START 3
    74 XXX
    75 XXX
    76 XXX
    77 OOO
    78 OOO
    79 OOO
    80 XXX
    81 XXX
    82 XXX
    83 0 0 1
    84 2 2 1
    85 END
    86 */
    ——现在的努力是为了小时候吹过的牛B!!
  • 相关阅读:
    视频直播:Windows中各类画面源的截取和合成方法总结
    短视频技术详解:Android端的短视频开发技术
    视频直播关键技术:流畅、拥塞和延时追赶
    视频直播技术详解:直播的推流调度
    直播技术:从性能参数到业务大数据,浅谈直播CDN服务监控
    音视频通话:小议音频处理与压缩技术
    pip命令报错“no perl script found in input”
    python常见面试题讲解(三)明明的随机数
    如何使用photoshop修改图片的像素大小(分辨率)
    VMware Workstation如何修改弹出释放快捷键
  • 原文地址:https://www.cnblogs.com/pingge/p/3125460.html
Copyright © 2011-2022 走看看