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  • Dungeon Master 和上题基本相同 bfs(),求最优值 不同的是三维bfs(),存储有技巧 (u=x*r*c+y*c+z);

    Problem Description
    You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides. 

    Is an escape possible? If yes, how long will it take? 
     
    Input
    The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). 
    L is the number of levels making up the dungeon. 
    R and C are the number of rows and columns making up the plan of each level. 
    Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.
     
    Output
    Each maze generates one line of output. If it is possible to reach the exit, print a line of the form 
    Escaped in x minute(s).

    where x is replaced by the shortest time it takes to escape. 
    If it is not possible to escape, print the line 
    Trapped!
     
    Sample Input
    3 4 5 S.... .###. .##.. ###.# ##### ##### ##.## ##... ##### ##### #.### ####E

     **************************************************************************************************************************

    bfs()  存储有技巧

    ***************************************************************************************************************************

     1 #include<iostream>
     2 #include<string>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<cstdio>
     6 #include<queue>
     7 using namespace std;
     8 int dx[6]={1,-1,0,0,0,0};
     9 int dy[6]={0,0,0,0,1,-1};
    10 int dz[6]={0,0,1,-1,0,0};
    11 int maze[101][101][101];
    12 int vis[101][101][101];
    13 int dis[101][101][101];
    14 int l,r,c,i,j,k;
    15 int que[100001];
    16 int stx,sty,stz;
    17 int bfs(int xt,int yt,int zt)
    18 {
    19     int rear,front;
    20     rear=front=0;
    21     int u=xt*r*c+yt*c+zt;//经典,使数存的更简洁
    22     que[rear++]=u;
    23     vis[xt][yt][zt]=1;
    24     while(front<rear)
    25     {
    26         int su=que[front];
    27         front++;
    28         int x=su/(r*c);
    29         int y=(su%(r*c))/c;
    30         int z=su%c;
    31         for(int it=0;it<6;it++)
    32         {
    33             int xa=x+dx[it];
    34             int ya=y+dy[it];
    35             int za=z+dz[it];
    36             if(xa>=0&&xa<l&&ya>=0&&ya<r&&za>=0&&za<c&&maze[xa][ya][za]!=0&&!vis[xa][ya][za])
    37             {
    38                 u=xa*r*c+ya*c+za;
    39                 que[rear++]=u;
    40                 dis[xa][ya][za]=dis[x][y][z]+1;
    41                 vis[xa][ya][za]=1;
    42                 if(maze[xa][ya][za]==2)
    43                  return  dis[xa][ya][za];
    44             }
    45         }
    46     }
    47     return 0;
    48 }
    49 int main()
    50 {
    51     char  ch;
    52     while(scanf("%d%d%d",&l,&r,&c)&&l&&r&&c)
    53     {
    54         getchar();
    55         memset(vis,0,sizeof(vis));
    56         memset(dis,0,sizeof(dis));
    57         memset(maze,0,sizeof(maze));
    58         for(i=0;i<l;i++)
    59         {
    60             for(j=0;j<r;j++)
    61             {
    62                 for(k=0;k<c;k++)
    63                 {
    64                     scanf("%c",&ch);
    65                     if(ch=='#')maze[i][j][k]=0;
    66                     if(ch=='.')maze[i][j][k]=1;
    67                     if(ch=='E')maze[i][j][k]=2;
    68                     if(ch=='S')
    69                     {
    70                         stx=i;
    71                         sty=j;
    72                         stz=k;
    73                     }
    74                 }
    75                 getchar();
    76             }
    77             getchar();
    78         }
    79     int ans=bfs(stx,sty,stz);
    80     if(ans==0)
    81       printf("Trapped!
    ");
    82     else
    83       printf("Escaped in %d minute(s).
    ",ans);
    84     }
    85     return 0;
    86 
    87 }
    View Code

    坚持!!!!!!!!!!!

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  • 原文地址:https://www.cnblogs.com/sdau--codeants/p/3369267.html
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