【NOI OJ】 2.5 搜索 1253:Dungeon Master（地牢逃脱）

描述

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? 

输入

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.

输出

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!

样例输入

3 4 5
S....
.###.
.##..
###.#

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

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

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

0 0 0

样例输出

Escaped in 11 minute(s).
Trapped!

   题目大意是：你被困在一个地牢里，现在正在S点，需要走到E点。迷宫有多层，多行，多列，可以向上走，向下走，或者前后左右，求最短路，如果无法到达的话输出无解。这是一道经典的走迷宫问题，只不过从二维迷宫变成了拥有层数的三维迷宫，30^3的数据范围，用一般的深度优先搜索枚举路线是根本不可能了（表示基本只会深搜），果然还是需要用最近才学的广搜勉强试试（汗）。感觉多了一维其实并没有变难多少，只是数组开成了三维，方向增加了上下，增加了一个层数的队列。但是我一开始的代码刚交上去就TLE了，果然广搜还是没有学透。附上超时的代码：

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m,l,sum;
char s[35][35][35];
int time[35][35][35];
int dir[6][3]={{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}};
queue<int>X;
queue<int>Y;
queue<int>Z;
bool check(int x,int y,int z)
{
if(x<0||y<0||z<0||x>=l||y>=n||z>=m||s[x][y][z]=='#')
{
return 0;
}
return 1;
}
void bfs()
{
while(!X.empty())
{
s[X.front()][Y.front()][Z.front()]='#';
for(int i=0;i<6;i++)
{
int x=X.front()+dir[i][0];
int y=Y.front()+dir[i][1];
int z=Z.front()+dir[i][2];
if(check(x,y,z)==1)
{
X.push(x);
Y.push(y);
Z.push(z);
time[x][y][z]=time[X.front()][Y.front()][Z.front()]+1;
if(s[x][y][z]=='E')
{
sum=time[x][y][z];
return ;
}
}
}
X.pop();
Y.pop();
Z.pop();
}
}
int main()
{
while(1)
{
scanf("%d %d %d\n",&l,&n,&m);
while(!X.empty())
{
X.pop();
Y.pop();
Z.pop();
}
if(l==0&&m==0&&n==0)
return 0;
for(int i=0;i<l;i++)
{
for(int j=0;j<n;j++)
{
for(int k=0;k<m;k++)
{
scanf("%c",&s[i][j][k]);
if(s[i][j][k]=='S')
{
X.push(i);
Y.push(j);
Z.push(k);
}
}
getchar();
}
getchar();
}
bfs();
if(sum!=0)
{
printf("Escaped in %d minute(s).\n",sum);
}
else
{
printf("Trapped!\n");
}
memset(s,0,sizeof(s));
memset(time,0,sizeof(time));
sum=0;
}
}

    检查了半天，原来是找到子结点时没有标记（#），改了之后就AC了~

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int n,m,l,sum;
char s[35][35][35];
int time[35][35][35];
int dir[6][3]={{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}};
queue<int>X;
queue<int>Y;
queue<int>Z;
bool check(int x,int y,int z)
{
if(x<0||y<0||z<0||x>=l||y>=n||z>=m||s[x][y][z]=='#')
{
return 0;
}
return 1;
}
void bfs()
{
while(!X.empty())
{
for(int i=0;i<6;i++)
{
int x=X.front()+dir[i][0];
int y=Y.front()+dir[i][1];
int z=Z.front()+dir[i][2];
if(check(x,y,z)==1)
{
X.push(x);
Y.push(y);
Z.push(z);
time[x][y][z]=time[X.front()][Y.front()][Z.front()]+1;
if(s[x][y][z]=='E')
{
sum=time[x][y][z];
return ;
}
s[x][y][z]='#';
}
}
X.pop();
Y.pop();
Z.pop();
}
}
int main()
{
while(1)
{
scanf("%d %d %d\n",&l,&n,&m);
while(!X.empty())
{
X.pop();
Y.pop();
Z.pop();
}
if(l==0&&m==0&&n==0)
return 0;
for(int i=0;i<l;i++)
{
for(int j=0;j<n;j++)
{
for(int k=0;k<m;k++)
{
scanf("%c",&s[i][j][k]);
if(s[i][j][k]=='S')
{
X.push(i);
Y.push(j);
Z.push(k);
s[i][j][k]='#';
}
}
getchar();
}
getchar();
}
bfs();
if(sum!=0)
{
printf("Escaped in %d minute(s).\n",sum);
}
else
{
printf("Trapped!\n");
}
memset(s,0,sizeof(s));
memset(time,0,sizeof(time));
sum=0;
}
}