[kuangbin带你飞]专题一简单搜索 B - Dungeon Master（POJ 2251）

最新推荐文章于 2020-02-07 16:24:45 发布

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最新推荐文章于 2020-02-07 16:24:45 发布

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本文介绍了一种解决三维迷宫问题的算法，利用BFS（宽度优先搜索）寻找从起点到终点的最短路径。文章详细解释了如何构建迷宫模型，并通过具体示例演示了算法的工作原理。

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B - Dungeon Master

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2251

Appoint description:

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：有一个三维地图长度分别为L,R,C， '.'这个为可通过区域，‘#’这个为墙， ‘S'这个为开始位置，’E'这个为出口，求从起点到出口的步数，如果不能到达输出Trapped!，否者输出步数。

求最短距离问题用bfs，bfs记得要判重。

详见代码。

简单搜索题考的是注意。

///注意细节就因为把=写成=好浪费一个小时
///当代码没有达到预期效果先看看逻辑上错没错， 然后再仔细查看代码
#include
  
   
#include
   
    
#include
    
     
#include
     
      
using namespace std;

const int N = 33;
char str[N][N][N];
int l, r, c;
int ans = 0;
struct point
{
    int x;
    int y;
    int z;
    int step;
    point(){}
    point(int xx, int yy, int zz, int ss)
    {
        x = xx;y = yy;z = zz;step = ss;
    }
};
queue
      
       que;
int dirx[6] = {1, -1, 0, 0, 0, 0};//用一些小技巧避免重复的代码，这样容易错
int diry[6] = {0, 0, 1, -1, 0, 0};
int dirz[6] = {0, 0, 0, 0, 1, -1};

int solve(point star)
{
    while(!que.empty()) que.pop();
    que.push(star);
    while(!que.empty())
    {
        point tmp = que.front();
        que.pop();
        for(int i = 0; i < 6; i++)
        {
            int x = tmp.x + dirx[i];
            int y = tmp.y + diry[i];
            int z = tmp.z + dirz[i];
            if(0 <= x && x < l && 0 <= y && y <= r && 0 <= z && z < c)
            {
                if(str[x][y][z] == 'E') {ans = tmp.step+1;return 1;}
                if(str[x][y][z] == '.')
                {
                    str[x][y][z] = '#';
                    que.push(point(x, y, z, tmp.step+1));
                }
            }
        }
    }
    return 0;
}

int main(void)
{

    while(scanf("%d%d%d", &l, &r, &c), l+r+c>0)
    {
        int i, j, k;
        point star;
        for(i = 0; i < l; i++)
            for(j = 0; j < r; j++)
            {
                scanf("%s", str[i][j]);
                for(k = 0; str[i][j][k]; k++)
                {
                    if(str[i][j][k] == 'S')
                    {
                        star.x = i;
                        star.y = j;
                        star.z = k;
                        star.step = 0;
                    }
                }
            }
        int tmp = solve(star);
        if(tmp == 0)
        {
            printf("Trapped!\n");
        }
        else
        {
            printf("Escaped in %d minute(s).\n", ans);
        }
    }
    return 0;
}