[UVA532] Dungeon Master 题解

Posted 2021-02-18 water-lift

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 file 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!

Solution

Different to traditional maze puzzles, it‘s a 3-dimensional maze game.

But actually they are the same. The only thing we have to do is turn 4-direction BFS into 6-direction one.

Implementation

Originlly, we have two array named dx and dy:

const int dx[4] = {1, 0, -1, 0},
          dy[4] = {0, 1, 0, -1};
// (dx[0], dy[0]) => right
// (dx[1], dy[1]) => forward
// (dx[2], dy[2]) => left
// (dx[3], dy[3]) => backward

When it comes to 3 dimensions, we have one dimension(z) and two direction(up&down) more.

const int dz[6] = {1, 0, 0, -1, 0, 0},
          dx[6] = {0, 1, 0, 0, -1, 0},
          dy[6] = {0, 0, 1, 0, 0, -1};
// (dx[0], dy[0], dz[0]) => up
// (dx[1], dy[1], dz[1]) => right
// (dx[2], dy[2], dz[2]) => forward
// (dx[3], dy[3], dz[3]) => down
// (dx[4], dy[4], dz[4]) => left
// (dx[5], dy[5], dz[5]) => backward

That‘s the core of this algorithm.

Chore

Input is terminated by three zeroes for L, R and C.

In fact, most problems on UVa are like this. You have to write a while(true) loop, read inputs at the beginning of the loop, and break until you‘ve got an invalid input.

Code

#include <bits/stdc++.h>
using namespace std;
int l, n, m;
struct pos
{
    int z, x, y, step;
    pos(int z, int x, int y, int step) : z(z), x(x), y(y), step(step) {}
};
int sz, sx, sy;
int tz, tx, ty;
bool mat[31][31][31];
bool vis[31][31][31];
const int dz[6] = {1, 0, 0, -1, 0, 0},
          dx[6] = {0, 1, 0, 0, -1, 0},
          dy[6] = {0, 0, 1, 0, 0, -1};
bool pan(int z, int x, int y)
{
    return z >= 1 && z <= l && x >= 1 && x <= n && y >= 1 && y <= m && mat[z][x][y] == 0;
}
int main()
{
    while (true)
    {
        cin >> l >> n >> m;
        if (l == 0 && n == 0 && m == 0)
            break;
        for (int k = 1; k <= l; k++)
        {
            for (int i = 1; i <= n; i++)
            {
                for (int j = 1; j <= m; j++)
                {
                    char c;
                    cin >> c;
                    if (c == ‘.‘)
                        mat[k][i][j] = 0;
                    else if (c == ‘#‘)
                        mat[k][i][j] = 1;
                    else if (c == ‘S‘)
                    {
                        mat[k][i][j] = 0;
                        sz = k;
                        sx = i;
                        sy = j;
                    }
                    else if (c == ‘E‘)
                    {
                        mat[k][i][j] = 0;
                        tz = k;
                        tx = i;
                        ty = j;
                    }
                    else
                    {
                        cin >> c;
                    }
                    vis[k][i][j] = 0;
                }
            }
        }
        queue<pos> q;
        q.push(pos(sz, sx, sy, 0));
        vis[sz][sx][sy] = 1;
        bool flag = 1;
        while (!q.empty())
        {
            pos head = q.front();
            q.pop();
            if (head.z == tz && head.x == tx && head.y == ty)
            {
                cout << "Escaped in " << head.step << " minute(s).
";
                flag = 0;
                break;
            }
            for (int i = 0; i < 6; i++)
            {
                int zz = head.z + dz[i], xx = head.x + dx[i], yy = head.y + dy[i];
                if (!pan(zz, xx, yy))
                    continue;
                if (vis[zz][xx][yy])
                    continue;
                q.push(pos(zz, xx, yy, head.step + 1));
                vis[zz][xx][yy] = 1;
            }
        }
        if (flag)
            cout << "Trapped!
";
    }
}