POJ 2251-Dungeon Master（BFS）
Dungeon Master
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 17007 | Accepted: 6620 |
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?
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.
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
where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line
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
3维地图给起点和终点搜最短路。。刷dfs专题刷出来这玩意也是醉了。。
BFS暴搜
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cctype>
#include <vector>
#include <cstdio>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define ll long long
#define maxn 360
#define pp pair<int,int>
#define INF 0x3f3f3f3f
#define max(x,y) ( ((x) > (y)) ? (x) : (y) )
#define min(x,y) ( ((x) > (y)) ? (y) : (x) )
using namespace std;
int n,m,sx,sy,sz,A,B,C;
bool vis[105][105][105];
char ma[105][105][105];
struct node
{
int x,y,z,step;
};
int mv[6][3]={{0,0,1},{0,0,-1},{1,0,0},{-1,0,0},{0,1,0},{0,-1,0}};
void bfs()
{
queue <node> Q;
node t;
t.x=sx;t.y=sy;t.z=sz;t.step=0;
vis[sx][sy][sz]=1;
Q.push(t);
while(!Q.empty())
{
node f=Q.front();Q.pop();
if(ma[f.x][f.y][f.z]=='E')
{
printf("Escaped in %d minute(s).\n",f.step);
return ;
}
for(int i=0;i<6;i++)
{
t.x=f.x+mv[i][0];
t.y=f.y+mv[i][1];
t.z=f.z+mv[i][2];
if(0<=t.x&&t.x<A&&0<=t.y&&t.y<B&&0<=t.z&&t.z<C&&!vis[t.x][t.y][t.z]&&ma[t.x][t.y][t.z]!='#')
{
vis[t.x][t.y][t.z]=1;
t.step=f.step+1;
Q.push(t);
}
}
}
puts("Trapped!");
}
int main()
{
while(scanf("%d%d%d",&A,&B,&C)!=EOF)
{
if(!A&&!B&&!C)break;
memset(vis,0,sizeof(vis));
for(int i=0;i<A;i++)
for(int j=0;j<B;j++)
scanf("%s",ma[i][j]);
for(int i=0;i<A;i++)
for(int j=0;j<B;j++)
for(int k=0;k<C;k++)
if(ma[i][j][k]=='S')
{
sx=i;sy=j;sz=k;
break;
}
bfs();
}
return 0;
}
