Dungeon Master POJ

对于3维的，可以用结构体来储存，详细见下列代码。

样例可以过，不过能不能ac还不知道，疑似poj炸了，

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
int dx[] = {1, -1, 0, 0, 0, 0};
int dy[] = {0, 0, -1, 1, 0, 0};
int dz[] = {0, 0, 0, 0, 1, -1};
int bx, by, bz;
int ex, ey, ez;
const int maxn = 30 + 5;
class POS
{
    public:
        POS():x_(-1), y_(-1), z_(-1){ }
        POS(int x, int y, int z):x_(x), y_(y), z_(z){ }
        int getX(){ return x_;}
        int getY(){ return y_;}
        int getZ(){ return z_;}
    private:
        int x_, y_, z_;
};
typedef POS P;
struct Level
{
    char maze[maxn][maxn];
    int vis[maxn][maxn];
    int d[maxn][maxn];
};
int L, R, C;
Level *h;
bool op(int x, int y, int z)
{
    if(x >= 0 && x < R && y >= 0 && y < C && z >= 0 && z < L)
        return true;
    return false;
}
int bfs()
{
    queue<P> pos;
    pos.push(P(bx, by, bz));
    while(!pos.empty())
    {
        P p = pos.front();
        pos.pop();
        if(p.getX() == ex && p.getY() == ey && p.getZ() == ez)
            break;
        h[p.getZ()].vis[p.getX()][p.getY()] = 1;
        for(int i = 0; i < 6; i++)
        {
            int curx = p.getX() + dx[i], cury = p.getY() + dy[i], curz = p.getZ() + dz[i];
            if(op(curx, cury, curz) && h[curz].maze[curx][cury] != '#' && !h[curz].vis[curx][cury])
            {
                h[curz].vis[curx][cury] = 1;
                pos.push(P(curx, cury, curz));
                h[curz].d[curx][cury] = h[p.getZ()].d[p.getX()][p.getY()] + 1;
            }
        }

    }
    return h[ez].d[ex][ey];

}
int main()
{
    while(cin >> L >> R >> C && (L || R || C))
    {
        bx = by = bz = -1;
        ex = ey = ez = -1;
        h = new Level[L];
        for(int i = 0; i < L; i++)
        {
            for(int j = 0; j < R; j++)
            {
                scanf("%s", h[i].maze[j]);      //input
                for(int k = 0; k < C ; k++)
                {
                    h[i].vis[j][k] = 0;
                    if(h[i].maze[j][k] == 'S')
                    {
                        h[i].d[j][k] = 0;
                        bx = j;
                        by = k;
                        bz = i;
                    }
                    else
                        h[i].d[j][k] = INF;
                    if(h[i].maze[j][k] == 'E')
                    {
                        ex = j;
                        ey = k;
                        ez = i;
                    }
                }

            }

        }
        int ans = bfs();
        if(ans == INF)
            cout << "Trapped!" << endl;
        else
            cout << "Escaped in " << ans << " minute(s)." << endl;




        delete []h;

    }

}