poj2251 三维bfs求最短距离

清疚 2022-08-18 11:56 192阅读 0赞

Dungeon Master

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong> 1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong> 65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong> 22466</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong> 8769</td> 
  </tr> 
 </tbody> 
</table>

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
    
    1 3 3
    S##
    #E#
    ###
    
    0 0 0

Sample Output

Escaped in 11 minute(s).
    Trapped!

题意：三维的地图，求起点到终点的最短距离

分析：三维bfs，多判断两个方向上下就行 。

#include<bitset>
    #include<map>
    #include<vector>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<queue>
    #include<set>
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;
        while((c=getchar())<'0' || c>'9');
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res;
    }
    struct st
    {
        int x,y,z;
    };
    const int N=100100;
    char a[33][33][33]; //层数 x y
    int dis[33][33][33];
    int l,n,m;
    int dx[]={-1,0,1,0},dy[]={0,-1,0,1};
    
    int bfs(st start,st end)
    {
        queue<st> q;
        q.push(start);
        dis[start.x][start.y][start.z]=0;
        while(!q.empty())
        {
            st now = q.front();
            q.pop();
            if(now.x==end.x && now.y==end.y && now.z==end.z) return dis[end.x][end.y][end.z];
    
            for(int i=0;i<4;i++)
            {
                int nk=now.x;
                int nx=now.y+dx[i];
                int ny=now.z+dy[i];
                if(nk>=0 && nx>=0 && ny>=0 && nk<l && nx<n && ny<m && a[nk][nx][ny]=='.' && dis[nk][nx][ny]==inf)
                {
                    dis[nk][nx][ny]=dis[now.x][now.y][now.z]+1;
                    q.push(st{nk,nx,ny});
                }
            }
            int nk=now.x+1;
            int nx=now.y;
            int ny=now.z;
            if(nk>=0 && nk<l && a[nk][nx][ny]=='.' && dis[nk][nx][ny]==inf)
            {
                dis[nk][nx][ny]=dis[now.x][now.y][now.z]+1;
                q.push(st{nk,nx,ny});
            }
            nk=now.x-1;
            nx=now.y;
            ny=now.z;
            if(nk>=0 && nk<l && a[nk][nx][ny]=='.' && dis[nk][nx][ny]==inf)
            {
                dis[nk][nx][ny]=dis[now.x][now.y][now.z]+1;
                q.push(st{nk,nx,ny});
            }
    
        }
        return dis[end.x][end.y][end.z];
    }
    int main()
    {
        while(~scanf("%d%d%d",&l,&n,&m) && l)
        {
            for(int k=0;k<l;k++)
            {
                for(int i=0;i<n;i++) scanf("%s",a[k][i]);
            }
            mem(dis,inf);
            st start,end;
            for(int k=0;k<l;k++)
            {
                for(int i=0;i<n;i++)
                {
                    for(int j=0;j<m;j++)
                    {
                        if(a[k][i][j]=='S')
                        {
                            start=st{k,i,j};
    
                        }
                        else if(a[k][i][j]=='E')
                        {
                            end=st{k,i,j};
                            a[k][i][j]='.';//注意要把end位置设置为'.',要不然会找不到，这里错了好几次
                        }
                    }
                }
            }
            int ans=bfs(start,end);
    
            if(ans==inf) puts("Trapped!");
            else
                printf("Escaped in %d minute(s).\n",ans);
        }
        return 0;
    }