POJ 2251 三維廣搜。
B - Dungeon Master
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Time Limit:1000MS
Memory Limit:65536KB
64bit IO Format:%I64d & %I64u
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Status
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!
作為一個弱菜，被這道題卡了一天了。
三維廣搜，六個方向，上下左右前後。
第一個案例可以理解為三個三層疊在一起的。
#include
#include
#include
#include
#include
using namespace std;
#define M 45
#define inf 0x6ffffff
char map[M][M][M];
int vis[M][M][M];
int dir[6][3]={{0,1,0},{0,-1,0},{1,0,0},{-1,0,0},{0,0,1},{0,0,-1}};//六個方向
int n,m,ok,k;
struct node
{
int x,y,z;
int time;
}
;
node f[666];
int ztime;
int z2,x2,y2,z1,x1,y1;
void bfs()
{
int i;
queueq;
node st,ed;
st.x=x1;
st.y=y1;
st.z=z1;
st.time=0;
q.push(st);
while(!q.empty())
{
st=q.front();
q.pop();
if(st.x==x2 &&st.y==y2 &&st.z==z2)
{
ok=1;
ztime=st.time;
return;
}
for(i=0;i<6;i++)
{
ed.x=st.x+dir[i][0];
ed.y=st.y+dir[i][1];
ed.z=st.z+dir[i][2];
if(map[ed.x][ed.y][ed.z]=='#' ||vis[ed.x][ed.y][ed.z] ||ed.x<0 ||ed.x>=k ||ed.y<0 ||ed.y>=n ||ed.z<0||ed.z>=m)//越界，搜過的地方，牆全部排除
continue;
ed.time=st.time+1; //時間加一
vis[ed.x][ed.y][ed.z]=1;
q.push(ed);
}
}
return;
}
int main()
{
int i,j,r;
while(scanf("%d%d%d",&k,&n,&m)!=EOF &&k!=0 &&n!=0 &&m!=0)
{
ok=0;
memset(vis,0,sizeof(vis));
for(i=0;i
