POJ 2251 Dungeon Master(BFS)

Posted 2020-09-27 ventricle

题目网址：http://poj.org/problem?id=2251

题目：

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34733		Accepted: 13268

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：

这道题无非就是将方向数组再增加一个维度——层数。由原来的四个方向，延伸为6个方向，多了上楼和下楼。总而言之就是换汤不换药，直接BFS。要注意的是每次搜索完都要清空一下队列。

代码：

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <queue>
 4 using namespace std;
 5 const int INF=111111111;
 6 struct node{
 7     int z,x,y;
 8 }dir[6]={{0,1,0},{0,-1,0},{0,0,1},{0,0,-1},{-1,0,0},{1,0,0}};
 9 queue<node>q;
10 int l,r,c;
11 int xb,yb,zb,xe,ye,ze;
12 int visited[35][35][35];
13 int flag;
14 char maze[35][35][35];
15 bool check(int z,int x,int y){
16     if(z<0 || z>=l) return false;
17     if(x<0 || x>=r) return false;
18     if(y<0 || y>=c) return false;
19     if(maze[z][x][y]==‘#‘)  return false;//这里的判断条件不要写成maze[z][x][y]!=‘.‘,因为还有结束标记‘E‘也是要走的
20     if(visited[z][x][y])    return false;//走过的点无需再走，因为再经过该点时说明耗费的时间已经比之前久了，所以不用考虑
21     return true;
22 }
23 void init(){
24     memset(visited, 0, sizeof(visited));
25     flag=0;
26     while (!q.empty())  q.pop();//记住每次清空队列
27 }
28 void bfs(){
29     node now,next;
30     now.z=zb;now.x=xb;now.y=yb;
31     q.push(now);
32     while (!q.empty()) {
33         now=q.front();q.pop();
34         if(now.z==ze && now.x==xe && now.y==ye){
35             flag=1;
36             printf("Escaped in %d minute(s).\n",visited[now.z][now.x][now.y]-1);//因为起点是从1开始标记，所以答案减一
37             break;
38         }
39         for(int d=0;d<6;d++){
40             next.z=now.z+dir[d].z;
41             next.x=now.x+dir[d].x;
42             next.y=now.y+dir[d].y;
43             if(check(next.z,next.x,next.y)){
44                 visited[next.z][next.x][next.y]=visited[now.z][now.x][now.y]+1;
45                 q.push(next);
46             }
47         }
48     }
49 }
50 int main(){
51     while(scanf("%d%d%d ",&l,&r,&c)!=EOF && (l+r+c)){
52         init();
53         for (int i=0; i<l; i++) {
54             for (int j=0; j<r; j++) {
55                 gets(maze[i][j]);
56                 for(int k=0; k<c; k++){
57                     if(maze[i][j][k]==‘S‘){
58                         zb=i;xb=j;yb=k;
59                     }
60                     if(maze[i][j][k]==‘E‘){
61                         ze=i;xe=j;ye=k;
62                     }
63                 }
64             }
65             char str[100];
66             gets(str);
67         }
68         visited[zb][xb][yb]=1;//起点要先标记走过
69         bfs();
70         if(!flag)   printf("Trapped!\n");
71     }
72     return 0;
73 }