POJ2251Dungeon Master(三维图)(BFS)

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17444		Accepted: 6790

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!

传送门

题意：某动物被困在三维的地下城里，起点是S，终点是E问他能不能逃跑，能逃最快需要几分钟。

思路：简单的bfs，三维就是二维的小升级，所以nxt数组要加一个上下移动的步骤，三维map即可

//272K	32MS
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int l,r,c,flag;
char mapp[35][35][35];
bool book[35][35][35];
struct point{
    int z;
    int x;
    int y;
    int t;
};
int nx[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}}; //加两个上下移动的操作
queue<point> que;
point s;
void ini()
{
    flag=0;
    memset(mapp,-1,sizeof(mapp));//把mapp初始化成-1，图的外界部分都是小于0的数所以判断边界就变得简单了
    memset(book,0,sizeof(book));
    for(int k=1;k<=l;k++){
        for(int i=1;i<=r;i++){
            for(int j=1;j<=c;j++){
                scanf("%c",&mapp[k][i][j]);
                if(mapp[k][i][j]=='S')
                s.z=k,s.x=i,s.y=j,s.t=0;
            }
            getchar(); //对这种字符的读入要格外注意换行符的读入问题
        }
        getchar();
    }
    while(!que.empty()) que.pop();

}
void bfs()
{
    que.push(s);
    book[s.z][s.x][s.y]=1;
    while(!que.empty()){
        point tmp=que.front();
        que.pop();
        if(mapp[tmp.z][tmp.x][tmp.y]=='E'){flag=1;printf("Escaped in %d minute(s).\n",tmp.t);return;}
        point nxt;int a,b,c;
        for(int i=0;i<6;i++){
            a=tmp.z+nx[i][0],b=tmp.x+nx[i][1],c=tmp.y+nx[i][2];
            if(mapp[a][b][c]<0||mapp[a][b][c]=='#'||book[a][b][c]) continue;
            nxt.z=a,nxt.x=b,nxt.y=c,nxt.t=tmp.t+1;
            que.push(nxt);
            book[a][b][c]=1;
        }
    }
}
int main()
{
    while(~scanf("%d%d%d",&l,&r,&c)&&(l||r||c)){
        getchar();
        ini();
        bfs();
        if(!flag) printf("Trapped!\n");
    }
    return 0;
}