poj 2251 Dungeon Master(三维BFS)(中等)

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20598		Accepted: 7971

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 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

思路：

搜索的地图是三维的，其他不变。直接BFS。

代码：

#include <iostream>
#include<cstring>
#include<queue>
#include<cstdio>
using namespace std;

typedef struct node
{
int x,y,z;
int lenth;
}node;
node sta;

int xx[]={1,-1,0,0,0,0};
int yy[]={0,0,0,0,-1,1};
int zz[]={0,0,-1,1,0,0};
bool vis[40][40][40];
char ma[40][40][40];
int l,r,c;

int bfs()
{
queue<node> Q;
int dx,dy,dz;
memset(vis,false,sizeof(vis));
sta.lenth=0;
Q.push(sta);
vis[sta.x][sta.y][sta.z]=true;
while(!Q.empty())
{
node cur=Q.front();
Q.pop();
for(int i=0;i<6;i++)
{
node next;
dx=cur.x+xx[i];
dy=cur.y+yy[i];
dz=cur.z+zz[i];
if(!vis[dx][dy][dz] && (ma[dx][dy][dz]=='.' || ma[dx][dy][dz]=='E') && dx>=0 && dx<l && dy>=0 && dy<r && dz>=0 && dz<c)
{
vis[dx][dy][dz]=true;
next.x=dx;
next.y=dy;
next.z=dz;
next.lenth=cur.lenth+1;
Q.push(next);
if(ma[dx][dy][dz]=='E')
return next.lenth;
}
}
}
return 0;
}

int main()
{
int i,j,k,key;
while(scanf("%d%d%d\n",&l,&r,&c))
{
if(l==0 && r==0 && c==0) break;

for(i=0;i<l;i++,getchar())
for(j=0;j<r;j++,getchar())
for(k=0;k<c;k++)
{
scanf("%c",&ma[i][j][k]);
if(ma[i][j][k]=='S') sta.x=i,sta.y=j,sta.z=k;
}
key=bfs();
if(key)
printf("Escaped in %d minute(s).\n",key);
else
printf("Trapped!\n");
}
return 0;
}

//数组，AC
#include <iostream>
#include<cstring>
#include<cstdio>
using namespace std;

struct q{
int x,y,z;
}q[30000];
int length[30000];
int xx[]={1,-1,0,0,0,0};
int yy[]={0,0,0,0,-1,1};
int zz[]={0,0,-1,1,0,0};
bool dis[40][40][40];
char ma[40][40][40];
int l,r,c,sx,sy,sz;

int bfs()
{
int rear,front,dx,dy,dz,i;
memset(dis,false,sizeof(dis));
memset(length,0,sizeof(length));

q[0].x=sx,q[0].y=sy,q[0].z=sz;
front=rear=0;
while(front<=rear)
{
for(i=0;i<6;i++)
{
dx=q[front].x+xx[i];
dy=q[front].y+yy[i];
dz=q[front].z+zz[i];
if(!dis[dx][dy][dz] && (ma[dx][dy][dz]=='.' || ma[dx][dy][dz]=='E') && dx>=0 && dx<l && dy>=0 && dy<r && dz>=0 && dz<c)
{
dis[dx][dy][dz]=true;
q[++rear].x=dx;
q[rear].y=dy;
q[rear].z=dz;
length[rear]=length[front]+1;
if(ma[dx][dy][dz]=='E')
return length[rear];
}
}
front++;
}
return 0;
}

int main()
{
int i,j,k,key;
while(scanf("%d%d%d\n",&l,&r,&c))
{
if(l==0 && r==0 && c==0) break;

for(i=0;i<l;i++,getchar())
for(j=0;j<r;j++,getchar())
for(k=0;k<c;k++)
{
scanf("%c",&ma[i][j][k]);
if(ma[i][j][k]=='S') sx=i,sy=j,sz=k;
}
key=bfs();
if(key)	printf("Escaped in %d minute(s).\n",key);
else printf("Trapped!\n");
}
return 0;
}