POJ 2251：Dungeon Master

Dungeon Master

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 20687 Accepted: 8004

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!

一个Test给你几个地图，给了你一个三维的模型，其实质就是之前人在格子里走路只能是上下左右四个方向，这回如果你在中间的地图里的话，又多了两个方向，就叫成前后吧，其实质是bfs，中规中矩的一个模式而已。

代码：

#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <string>#include <cstring>#include <queue>using namespace std;int L,R,C;char value[50][50][50];int bushu[50][50][50];int bfs(int i,int j,int k){int flag=0;queue<int>x;queue<int>y;queue<int>z;x.push(i);y.push(j);z.push(k);while(x.size()){int temp1=x.front();int temp2=y.front();int temp3=z.front();x.pop();y.pop();z.pop();if(temp1-1>=1&&value[temp1-1][temp2][temp3]=='.'&&!bushu[temp1-1][temp2][temp3]){x.push(temp1-1);y.push(temp2);z.push(temp3);bushu[temp1-1][temp2][temp3]=bushu[temp1][temp2][temp3]+1;}if(temp1+1<=L&&value[temp1+1][temp2][temp3]=='.'&&!bushu[temp1+1][temp2][temp3]){x.push(temp1+1);y.push(temp2);z.push(temp3);bushu[temp1+1][temp2][temp3]=bushu[temp1][temp2][temp3]+1;}if(temp2-1>=1&&value[temp1][temp2-1][temp3]=='.'&&!bushu[temp1][temp2-1][temp3]){x.push(temp1);y.push(temp2-1);z.push(temp3);bushu[temp1][temp2-1][temp3]=bushu[temp1][temp2][temp3]+1;}if(temp2+1<=R&&value[temp1][temp2+1][temp3]=='.'&&!bushu[temp1][temp2+1][temp3]){x.push(temp1);y.push(temp2+1);z.push(temp3);bushu[temp1][temp2+1][temp3]=bushu[temp1][temp2][temp3]+1;}if(temp3-1>=1&&value[temp1][temp2][temp3-1]=='.'&&!bushu[temp1][temp2][temp3-1]){x.push(temp1);y.push(temp2);z.push(temp3-1);bushu[temp1][temp2][temp3-1]=bushu[temp1][temp2][temp3]+1;}if(temp3+1<=C&&value[temp1][temp2][temp3+1]=='.'&&!bushu[temp1][temp2][temp3+1]){x.push(temp1);y.push(temp2);z.push(temp3+1);bushu[temp1][temp2][temp3+1]=bushu[temp1][temp2][temp3]+1;}if(value[temp1-1][temp2][temp3]=='E'||value[temp1+1][temp2][temp3]=='E'||value[temp1][temp2-1][temp3]=='E'||value[temp1][temp2+1][temp3]=='E'||value[temp1][temp2][temp3-1]=='E'||value[temp1][temp2][temp3+1]=='E'){flag=1;return bushu[temp1][temp2][temp3]+1;}}return 0;}void solve(){int i,j,k;for(i=L;i>=1;i--){for(j=R;j>=1;j--){for(k=C;k>=1;k--){if(value[i][j][k]=='S'){int temp=bfs(i,j,k);if(temp==0)cout<<"Trapped!"<<endl;elsecout<<"Escaped in "<<temp<<" minute(s)."<<endl;return;}}}}}int main(){int i,j,k;while(scanf("%d%d%d",&L,&R,&C)){if(L+R+C==0)break;for(i=1;i<=L;i++){for(j=1;j<=R;j++){scanf("%s",value[i][j]+1);}}memset(bushu,0,sizeof(bushu));solve();}return 0;}