POJ 2251:dungeon Master "BFS"

Dungeon Mastertime limit:2000/1000ms (java/other) Memory limit:131072/65536k (Java/other) total submission (s): 9 A ccepted Submission (s): 7Problem descriptionyou is trapped in a 3D dungeon and need to find the quickest a- The dungeon is composed of a unit cubes which may or may isn't being 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 are surrounded by solid rock in all sides.

is an escape possible? If Yes, how long would it take?

Inputthe input consists of a number of dungeons. Each of the dungeon description starts with a line containing three integers L, R and C (all limited to the size).
L is the number of levels making up the dungeon.
R and C is the number of rows and columns making up the plan of each level.
Then there'll follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock was indicated by a ' # ' and empty cells were 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.
Outputeach 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 isn't possible to escape, print the line

trapped!

Ac-code:

#include <cstdio> #include <cstring> #include <queue> #define INF 0xffffffint ans,l,r,c,ex,ey,ez,vis[ 35][35][35];char a[35][35][35];int dx[6]={0,0,0,0,1,-1};int dy[6]={1,-1,0,0,0,0};int dz[6]={0,0,1,-1,0,0};using namespace std;struct Node {int x,y,z,step;friend bool operator < (node A,node b) {return a.step>b.step;}} temp,p;void BFS (int x1,int y1,int z1) {P.x=x1;p.y=y1;p.z=z1;p.step=0;vis[p.x][p.y][p.z]=1;priority_queue<node >q;q.push (P); while (!q.empty ()) {p=q.top (), Q.pop (); for (int i=0;i<6;i++) {temp.x=p.x+dx[i];temp.y=p.y+dy[i]; Temp.z=p.z+dz[i];temp.step=p.step+1;if (!vis[temp.x][temp.y][temp.z]&&temp.x>=0&&temp.x<l &AMP;&AMP;TEMP.Y&GT;=0&AMP;&AMP;TEMP.Y&LT;R&AMP;&AMP;TEMP.Z&GT;=0&AMP;&AMP;TEMP.Z&LT;C) {if (temp.x==ex& &temp.y==ey&&temp.z==ez) {printf ("escaped in%d minute (s). \ n", temp.step); return;} Vis[temp.x][temp.y][temp.z]=1;q.push (temp);}}} printf ("trapped!\n");} int main () {int i,j,k,x,y,z;while (scanf ("%d%d%d",&l,&r,&c), l| | r| | c) {memset (vis,0,sizeof (VIS)), for (I=0;i<l;i++,getchar ())} {for (j=0;j<r;j++) {GetChar (), for (k=0;k<c;k++) { scanf ("%c", &a[i][j][k]); if (a[i][j][k]== ' S ') {x=i;y=j;z=k;} else if (a[i][j][k]== ' E ') {ex=i;ey=j;ez=k;} else if (a[i][j][k]== ' # ') Vis[i][j][k]=1;elsevis[i][j][k]=0;}}} Ans=inf;bfs (x, y, z);} return 0;}

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POJ 2251:dungeon Master "BFS"