本文主要是介绍poj 2251Dungeon Master(BFS, DFS亦可)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
今天，一看题，我就想到用BFS，毕竟BFS是具有贪心性质的，可以用来求最小短的……给我的感觉就是简单，但是很烦，因为它是三维的，和poj3728 catch that cow差不多,需要一个数组来存储到达那一点的最小的时间;需要注意的是，输出‘.’;
Dungeon Master
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9761 | Accepted: 3777 |
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.... .###. .##.. ###.###### ##### ##.## ##...##### ##### #.### ####E1 3 3 S## #E# ###0 0 0
Sample Output
Escaped in 11 minute(s). Trapped!
#include<stdio.h>
#include<string.h>
struct Node
{int x;int y;int z;
}node[170000];
int l, r, c;char map[31][31][31];
int visited[31][31][31];
int step[31][31][31];int x1, y1, z1;
int x2, y2, z2;
int a[6][3]={{-1,0,0},{1,0,0},{0,-1,0},{0,1,0},{0,0,-1},{0,0,1}};int BFS(int x, int y, int z)
{int i;int X, Y, Z;int front=0, rear=1;node[front].x=x;node[front].y=y;node[front].z=z;visited[z][y][x]=1;while(front != rear){for(i=0; i<6; i++){X=node[front].x+a[i][0];Y=node[front].y+a[i][1];Z=node[front].z+a[i][2];if(X>=0&&X<c && Y>=0&&Y<r && Z>=0&&Z<l && !visited[Z][Y][X] && map[Z][Y][X]!='#'){step[Z][Y][X]=step[node[front].z][node[front].y][node[front].x]+1;if( X==x2 && Y==y2 && Z==z2){return 1;}node[rear].x=X;node[rear].y=Y;node[rear].z=Z;visited[Z][Y][X]=1;rear++;}}front++; }return 0;
}
int main()
{int i, j, k;while(scanf("%d %d %d", &l, &r, &c) && l){memset(visited, 0, sizeof(visited) );memset(step, 0, sizeof(step) );for(i=0; i<l; i++){for(j=0; j<r; j++){getchar();for(k=0; k<c; k++){scanf("%c", &map[i][j][k]);if(map[i][j][k]=='S'){z1=i;y1=j;x1=k;}else if(map[i][j][k]=='E'){z2=i;y2=j;x2=k;}else if(map[i][j][k]=='#')visited[i][j][k]=1;}}getchar();}if( BFS(x1, y1, z1) )printf("Escaped in %d minute(s).\n", step[z2][y2][x2]);elseprintf("Trapped!\n");}
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