POJ 2251-Dungeon Master (BFS)

POJ 2251-Dungeon Master (BFS)
Dungeon Master

Time Limit:1000 MS		Memory Limit:65536 K
Total Submissions:17007		Accepted:6620

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 '. '. 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 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0

The 3-dimensional map returns the shortest path to the start point and the end point .. I'm also drunk when I click the dfs topic ..

BFS Brute Force Search

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               #define INF 0x3f3f3f3f#define max(x,y) ( ((x) > (y)) ? (x) : (y) )#define min(x,y) ( ((x) > (y)) ? (y) : (x) )using namespace std;int n,m,sx,sy,sz,A,B,C;bool vis[105][105][105];char ma[105][105][105];struct node{ int x,y,z,step;};int mv[6][3]={{0,0,1},{0,0,-1},{1,0,0},{-1,0,0},{0,1,0},{0,-1,0}};void bfs(){ queue 
               
                 Q; node t; t.x=sx;t.y=sy;t.z=sz;t.step=0; vis[sx][sy][sz]=1; Q.push(t); while(!Q.empty()) { node f=Q.front();Q.pop(); if(ma[f.x][f.y][f.z]=='E') {printf("Escaped in %d minute(s).\n",f.step);return ; } for(int i=0;i<6;i++) { t.x=f.x+mv[i][0]; t.y=f.y+mv[i][1]; t.z=f.z+mv[i][2]; if(0<=t.x&&t.x