算法训练之BFS POJ 2251 Dungeon Master 三维最短路径

Dungeon Master

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 27195 Accepted: 10668

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!

Source

Ulm Local 1997

典型的最短路径问题，将二维的BFS扩展到三维即可。

在此给出AC代码：

//三维bfs #include<stdio.h>#include<iostream>#include<queue>using namespace std;const int maxn=31;int map[maxn][maxn][maxn];int vis[maxn][maxn][maxn];int sx,sy,sz,ex,ey,ez;int c,a,b;int dirx[6]= {-1,1,0,0,0,0};int diry[6]= {0,0,-1,1,0,0};int dirz[6]= {0,0,0,0,-1,1};structnode{int x,y,z,step;};int BFS(){queue<node>q;node p1;p1.x=sx;p1.y=sy;p1.z=sz;vis[sx][sy][sz]=1;p1.step=0;q.push(p1);//printf("%d %d %d\n",p1.x,p1.y,p1.z);while(!q.empty()){node f=q.front(),temp;q.pop();if(f.x==ex&&f.y==ey&&f.z==ez)return f.step;for(int i=0; i<6; i++){temp.x=f.x+dirx[i];temp.y=f.y+diry[i];temp.z=f.z+dirz[i];temp.step=f.step+1;if(temp.x<a&&temp.x>=0&&temp.y<b&&temp.y>=0&&temp.z<c&&temp.z>=0&&        !vis[temp.x][temp.y][temp.z]&&(map[temp.x][temp.y][temp.z]==1)){q.push(temp);//printf("%d %d %d\n",temp.x,temp.y,temp.z);vis[temp.x][temp.y][temp.z]=1;}}}return -1;}int main(){while(~scanf("%d%d%d",&c,&a,&b),a,b,c){char temp;for(int i=0; i<c; i++)//平面坐标for(int j=0; j<a; j++)//j是平面xfor(int k=0; k<b; k++)//k是平面y{vis[j][k][i]=0;cin>>temp;if(temp=='.')map[j][k][i]=1;else if(temp=='#')map[j][k][i]=0;else if(temp=='S'){sx=j;sy=k;sz=i;map[j][k][i]=1;}else if(temp=='E'){ex=j;ey=k;ez=i;map[j][k][i]=1;}}int  res=BFS();if(res==-1)printf("Trapped!\n");elseprintf("Escaped in %d minute(s).\n",res);}return 0;}