POJ-2251 Dungeon Master (搜索)-优快云博客

本文链接：https://blog.youkuaiyun.com/Lin_ZR/article/details/79112149

本文介绍了一种解决三维迷宫问题的算法，利用广度优先搜索(BFS)找到从起点到终点的最短路径。提供了两种实现方式的AC代码，并详细解释了代码的结构与逻辑。

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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?

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!

题意：给一个3D 的地牢，每一步的可以走东南西北上下6个方向，# 不能走，S是起点，E是终点，如果能走到终点就输出最短步数，如果不能就输出 Trapped!

思路：简单的BFS 在多加两个方向；

AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;

#define INF 0x7fffffff
struct Node{
    int n,l,r;
    Node(){}
    Node(int cn,int cl,int cr){
        n = cn;
        r = cr;
        l = cl;
    }
};
int n,l,r;
int sn,sl,sr,en,el,er;
int dx[6] = { -1, 1, 0, 0, 0, 0 };
int dy[6] = { 0, 0, -1, 1, 0, 0 };
int dz[6] = { 0, 0, 0, 0, 1, -1 };
char a[35][35][35];
int d[35][35][35];
int bfs(int sn,int sl,int sr){
    queue<Node> que;
    que.push(Node(sn,sl,sr));
	d[sn][sl][sr] = 0;
	while (que.size()){
		Node cur = que.front(); que.pop();
		//printf("%d\n", que.size());
		if (cur.l == el && cur.n == en && cur.r == r)
			break;
		for (int i = 0; i < 6; i++){
			int nn, nl, nr;
			nl = cur.l + dx[i];
			nr = cur.r + dy[i];
			nn = cur.n + dz[i];
			if (nn >= 0 && nn < n && nl < l && nr < r && nl >= 0 && nr >= 0 && d[nn][nl][nr] == INF && a[nn][nl][nr] != '#'){
				que.push(Node(nn, nl, nr));
				d[nn][nl][nr] = d[cur.n][cur.l][cur.r] + 1;
			}
		}
	}
    return d[en][el][er];
}
int main(){
#ifdef LOCAL
	freopen("Text.txt", "r", stdin);
#endif
    while(scanf("%d%d%d",&n,&l,&r)==3&&n&&l&&r){
        memset(d,0,sizeof(d));
        for(int i = 0; i < n; i++){
            for(int j = 0; j < l; j++){
                scanf("%s",a[i][j]);
                for(int k = 0; k < r; k++){
                    if(a[i][j][k] == 'S'){
                        sn = i;
                        sl = j;
                        sr = k;
                    }
                    if(a[i][j][k] == 'E'){
                        en = i;
                        el = j;
                        er = k;
                    }
					d[i][j][k] = INF;
                }
            }
        }
        int cnt = bfs(sn,sl,sr);
        if(cnt != INF)
            printf("Escaped in %d minute(s).\n",cnt);
        else
            printf("Trapped!\n");
    }
    return 0;
}

第二种：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;

#define INF 0x7fffffff
struct Node{
    int n,l,r;
    Node(){}
    Node(int cn,int cl,int cr){
        n = cn;
        r = cr;
        l = cl;
    }
};
int n,l,r;
int sn,sl,sr,en,el,er;
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
char a[35][35][35];
int d[35][35][35];
int bfs(int sn,int sl,int sr){
    queue<Node> que;
    que.push(Node(sn,sl,sr));
    int ans = 0;
	d[sn][sl][sr] = 0;
    while(que.size()){
        Node cur = que.front();que.pop();
		//printf("%d\n", que.size());
		if (cur.l == el && cur.n == en && cur.r == r)
            break;
        for(int i = 0; i < 4; i++){
            int nn,nl,nr;
            nl = cur.l + dx[i];
            nr = cur.r + dy[i];
            nn = cur.n;
			if (a[nn][nl][nr] != '#' && nl < l && nr < r && nl >= 0 && nr >= 0 && d[nn][nl][nr] == INF){
                que.push(Node(nn,nl,nr));
                d[nn][nl][nr] = d[cur.n][cur.l][cur.r] + 1;
            }
        }
		if (cur.n + 1 < n && a[cur.n + 1][cur.l][cur.r] != '#' && d[cur.n + 1][cur.l][cur.r] == INF){
             que.push(Node(cur.n + 1,cur.l,cur.r));
             d[cur.n + 1][cur.l][cur.r] = d[cur.n][cur.l][cur.r] + 1;
        }
		if (cur.n - 1 >= 0 && a[cur.n - 1][cur.l][cur.r] != '#' && d[cur.n - 1][cur.l][cur.r] == INF){
             que.push(Node(cur.n - 1,cur.l,cur.r));
             d[cur.n - 1][cur.l][cur.r] = d[cur.n][cur.l][cur.r] + 1;
        }
    }
    return d[en][el][er];
}
int main(){
#ifdef LOCAL
	freopen("Text.txt", "r", stdin);
#endif
    while(scanf("%d%d%d",&n,&l,&r)==3&&n&&l&&r){
        memset(d,0,sizeof(d));
        for(int i = 0; i < n; i++){
            for(int j = 0; j < l; j++){
                scanf("%s",a[i][j]);
                for(int k = 0; k < r; k++){
                    if(a[i][j][k] == 'S'){
                        sn = i;
                        sl = j;
                        sr = k;
                    }
                    if(a[i][j][k] == 'E'){
                        en = i;
                        el = j;
                        er = k;
                    }
					d[i][j][k] = INF;
                }
            }
        }
        int cnt = bfs(sn,sl,sr);
        if(cnt != INF)
            printf("Escaped in %d minute(s).\n",cnt);
        else
            printf("Trapped!\n");

    }
    return 0;
}