POJ:Dungeon Master(三维bfs模板题）

最新推荐文章于 2020-02-08 00:35:09 发布

转载最新推荐文章于 2020-02-08 00:35:09 发布 · 102 阅读

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原文链接：http://www.cnblogs.com/zhangmingcheng/p/3973598.html

本文介绍了一种解决3D迷宫逃脱问题的算法。通过输入描述迷宫的三维矩阵，算法能判断是否存在从起点到终点的路径，并计算最短逃脱时间。详细解释了BFS搜索算法的应用，包括初始化状态、迭代更新以及最终结果输出。

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16748		Accepted: 6522

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!
题解：
就是很裸的模板题，细心就好。其中'S'是起点，‘E’是终点，‘#’不能走。

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
char map[51][51][51];
int fx[6]= {1,-1,0,0,0,0};
int fy[6]= {0,0,1,-1,0,0};
int fz[6]= {0,0,0,0,1,-1};
struct node
{
    int ans;
    int x,y,z;
};
struct node t,f;
int v[51][51][51];
int n,m,k;
void bfs(int xx,int yy,int zz)
{
    memset(v,0,sizeof(v));
    queue<node>q;
    t.x=xx;
    t.y=yy;
    t.z=zz;
    t.ans=0;
    q.push(t);
    v[t.x][t.y][t.z]=1;
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        if(map[t.x][t.y][t.z]=='E')
        {
            printf("Escaped in %d minute(s).\n",t.ans);
            return ;
        }
        for(int i=0; i<6; i++)
        {
            f.x=t.x+fx[i];
            f.y=t.y+fy[i];
            f.z=t.z+fz[i];
            if(f.x>=0&&f.x<n&&f.y>=0&&f.y<m&&f.z>=0&&f.z<k&&v[f.x][f.y][f.z]==0&&map[f.x][f.y][f.z]!='#')
            {
                v[f.x][f.y][f.z]=1;
                f.ans=t.ans+1;
                q.push(f);
            }
        }
    }
    printf("Trapped!\n");
    return ;
}
int main()
{
    int xx,yy,zz;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        if(n==0&&m==0&&k==0) break;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                scanf("%*c%s",map[i][j]);
            }
        }
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                for(int z1=0; z1<k; z1++)
                {
                    if(map[i][j][z1]=='S')
                    {
                        xx=i;
                        yy=j;
                        zz=z1;
                        break;
                    }
                }
            }
        }
        bfs(xx,yy,zz);
    }
    return 0;
}

转载于:https://www.cnblogs.com/zhangmingcheng/p/3973598.html