POJ - 2251 Dungeon Master 多维多方向BFS

Dungeon Master

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!之前做过类似的A计划，那道题是二维传送地图，当时是用DFS写的。这次的Dungeon Master是A计划的升级版，达到30的多维地图（WA：只能传送到相邻维度。。），所以尝试DFS果断超时，，然后重码了一遍BFS-_-16msA掉。这再一次地证明了在处理最短路径时BFS的效率之高。DFS TLE代码： #include<stdio.h> #include<string.h> char a[35][35][35]; int b[35][35][35]; int t[6][3]={{0,1,0},{0,0,1},{0,-1,0},{0,0,-1},{1,0,0},{-1,0,0}}; int ww,n,m,bw,bx,by,min; void dfs(int w,int x,int y,int s) { int i; if(w<0||x<0||y<0||w>=ww||x>=n||y>=m) return; if(a[w][x][y]=='E'){ if(s<min) min=s; return; } if(a[w][x][y]=='#'||b[w][x][y]==1) return; for(i=0;i<6;i++){ int tw=w+t[i][0]; int tx=x+t[i][1]; int ty=y+t[i][2]; b[w][x][y]=1; dfs(tw,tx,ty,s+1); b[w][x][y]=0; } } int main() { int i,j,k; while(scanf("%d%d%d",&ww,&n,&m)&&!(ww==0&&n==0&&m==0)){ for(i=0;i<ww;i++){ for(j=0;j<n;j++){ getchar(); scanf("%s",a[i][j]); for(k=0;k<m;k++){ if(a[i][j][k]=='S'){ bw=i; bx=j; by=k; } } } } min=1000000000; memset(b,0,sizeof(b)); dfs(bw,bx,by,0); if(min==1000000000) printf("Trapped!\n"); else printf("Escaped in %d minute(s).\n",min); } return 0; }

BFS AC代码：

#include<stdio.h> #include<string.h> #include<queue> using namespace std; char a[35][35][35]; int b[35][35][35]; int t[6][3]={{0,1,0},{0,0,1},{0,-1,0},{0,0,-1},{1,0,0},{-1,0,0}}; struct Node{ int w,x,y,s; }node; int main() { int ww,n,m,i,j,k; while(scanf("%d%d%d",&ww,&n,&m)&&!(ww==0&&n==0&&m==0)){ queue<Node> q; memset(b,0,sizeof(b)); for(i=0;i<ww;i++){ for(j=0;j<n;j++){ getchar(); scanf("%s",a[i][j]); for(k=0;k<m;k++){ if(a[i][j][k]=='S'){ b[i][j][k]=1; node.w=i; node.x=j; node.y=k; node.s=0; q.push(node); } } } } int f=0; while(q.size()){ for(i=0;i<6;i++){ int tw=q.front().w+t[i][0]; int tx=q.front().x+t[i][1]; int ty=q.front().y+t[i][2]; if(tw<0||tx<0||ty<0||tw>=ww||tx>=n||ty>=m) continue; if(a[tw][tx][ty]=='E'){ f=q.front().s+1; break; } if(a[tw][tx][ty]=='#'||b[tw][tx][ty]==1) continue; b[tw][tx][ty]=1; node.w=tw; node.x=tx; node.y=ty; node.s=q.front().s+1; q.push(node); } if(f!=0) break; q.pop(); } if(f==0) printf("Trapped!\n"); else printf("Escaped in %d minute(s).\n",f); } return 0; }

 

转载于:https://www.cnblogs.com/yzm10/p/7240302.html