2251 Dungeon Master【bfs】

Dungeon Master

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20969 Accepted: 8136

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

Sample Output

Escaped in 11 minute(s).Trapped!

bfs 题目，这个题看起来比较恐怖，因为是三维立体的，但是也不算复杂，因为在这个图里，没有特别复杂的限定条件，到达下一步的时间都一样，这样只需要简单的处理就可以了，不过注意标记，以及入队元素的判断，

bfs 优先本层搜索，之后在进行更深一层的搜索...

#include<stdio.h>#include<queue>#include<string.h>using namespace std;char x[50][50][50];int l,r,c,v[50][50][50],kase;int fx[4]={0,-1,1,0},fy[4]={-1,0,0,1},fz[2]={-1,1};//这样定义数组有利于后面操作struct ex{int x,y,z;//三个坐标int t;//用的时间}w,tp;queue<ex> q;void bfs(){while(!q.empty()){w=q.front();q.pop();if(x[w.x][w.y][w.z]=='E')//到达出口{kase=1;return;}for(int i=0;i<2;++i)//这里考虑的是上下两个方向{int tx=w.x,ty=w.y,tz=w.z+fz[i];tp=w;if(tz<0||tz>=c||x[tx][ty][tz]=='#'){continue;}if(!v[tx][ty][tz])//判断是否要入队{tp.z=tz;++tp.t;q.push(tp);v[tx][ty][tz]=1;}}for(int i=0;i<4;++i)//这里考虑前后左右的方向.{int tx=w.x+fx[i],ty=w.y+fy[i],tz=w.z;tp=w;if(tx<0||tx>=l||ty<0||ty>=r||x[tx][ty][tz]=='#')//越界或者不能行走{continue;}if(!v[tx][ty][tz])//判断状态是否出现过{tp.x=tx;tp.y=ty;++tp.t;q.push(tp);v[tx][ty][tz]=1;}}}}int main(){int i,j,k,a,b,d;char s;while(scanf("%d%d%d",&l,&r,&c),l|r|c){for(i=0;i<l;++i){for(j=0;j<r;++j)//输入要注意，格式的控制{getchar();for(k=0;k<c;++k){scanf("%c",&s);x[i][j][k]=s;if(s=='S'){a=i;b=j;d=k;}}}getchar();}memset(v,0,sizeof(v));while(!q.empty())//注意清空队列{q.pop();}w.x=a;w.y=b;w.z=d;w.t=0;v[a][b][c]=1;kase=0;q.push(w);//入队第一个元素bfs();if(kase)//满足条件{printf("Escaped in %d minute(s).\n",w.t);continue;}printf("Trapped!\n");}return 0;}