POJ - 2251 Dungeon Master (简单BFS)
Dungeon Master
| Time Limit:1000MS | Memory Limit:65536KB | 64bit IO Format:%I64d & %I64u |
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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?
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
Sample Output
Escaped in 11 minute(s). Trapped!
题目大意：
给出一三维空间的地牢,要求求出由字符'S'到字符'E'的最短路径
移动方向可以是上，下，左，右，前，后，六个方向
每移动一次就耗费一分钟，要求输出最快的走出时间。
不同L层的地图，相同RC坐标处是连通的。
其中'S'是入口，'E'是出口，'#'是墙壁，'.'是通路。
解析：求最短路问题，直接BFS
另外附上一组样例：
input:
3 4 5 S.... .###. .##.. ###.# ##### ##### ##.## ##... ##### ##### #.### ####E 1 3 3 S## #E# ### 4 4 5 ##### ##### ##.## ##... ##### ###S# #.#.# #.... ##### ##### #.### ####E ##### #...# ...## #.... 1 2 2 ## SE 3 3 3 ### #E# #.# #.. .## .#. ... ... ..S 5 2 2 ## ## #. .. .S .. .. .. .E .. 2 1 1 S E 1 1 2 SE 1 5 5 ##### S#### ....# ####E ##... 2 3 3 #.E ... ... ... ... ..S 2 4 4 #### #S## .... ###. #### .### .### .E.. 2 5 5 ##### #S..# ..#.. .###. .###. ##### ##### ##### ##### ..E.. 0 0 0
output:
Escaped in 11 minute(s). Trapped! Escaped in 4 minute(s). Escaped in 1 minute(s). Trapped! Escaped in 2 minute(s). Escaped in 1 minute(s). Escaped in 1 minute(s). Trapped! Escaped in 3 minute(s). Escaped in 5 minute(s). Escaped in 7 minute(s).
#include <iostream>
#include <queue>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N = 31;
const int dx[6]={1,-1,0,0,0,0};
const int dy[6]={0,0,1,-1,0,0};
const int dz[6]={0,0,0,0,1,-1};
char grid[N][N][N];
struct Node{
int x,y,z,step;
};
int l,r,c;
Node start,end;
int bfs(Node s) {
grid[s.x][s.y][s.z] = '#';
if(s.x==end.x && s.x==end.y && s.z==end.z) {
return s.step;
}
queue<Node> Q;
Q.push(s);
Node tmp;
while(!Q.empty()) {
Node now;
now = Q.front();
Q.pop();
for(int i = 0; i < 6; i++) {
tmp.x = now.x + dx[i];
tmp.y = now.y + dy[i];
tmp.z = now.z + dz[i];
tmp.step = now.step + 1;
if(tmp.x==end.x && tmp.y==end.y && tmp.z==end.z) {
return tmp.step;
}
if(tmp.x>=0 && tmp.x<l && tmp.y>=0 && tmp.y<r && tmp.z>=0 && tmp.z<c && grid[tmp.x][tmp.y][tmp.z]=='.')
{
grid[tmp.x][tmp.y][tmp.z]='#';
Q.push(tmp);
}
}
}
return -1;
}
int main() {
while(~scanf("%d%d%d",&l,&r,&c) && (l || r || c)) {
for(int i=0;i<l;i++) {
for(int j=0;j<r;j++) {
scanf("%s",grid[i][j]);
for(int k=0;k<c;k++) {
if( grid[i][j][k]=='S') {
start.x=i;
start.y=j;
start.z=k;
start.step = 0;
}
if( grid[i][j][k]=='E') {
end.x=i;
end.y=j;
end.z=k;
}
}
}
}
int ans;
ans = bfs(start);
if(ans!=-1)
printf("Escaped in %d minute(s).\n",ans);
else
printf("Trapped!\n");
memset(grid,0,sizeof(grid));
}
return 0;
}