POJ 2251 B - Dungeon Master——双广度优先搜索

题意：在一个三维空间中寻找起止点之间的最短路，每次可以朝着同一个平面的前后左右走，或者到上一个平面或下一个平面与当前平面相同的位置。

思路：bfs暴一遍超时，考虑双向bfs

#include 
#include 
#include 
#include 
#include 

using namespace std;

const int dx[] = {0, 0, -1, 1};
const int dy[] = {1, -1, 0, 0};
int l, r, c, ans, vis[50][50][50], visr[50][50][50];
struct Graph
{
    char g[50][50];
}G[50];
struct Point
{
    int x, y, z;
}st, ed;
struct State
{
    int x, y,z, step;
}state, temp;

void bfs()
{
    ans = -1;
    memset(vis, 0, sizeof(vis));
    memset(visr, 0, sizeof(visr));
    queue q;
    queue qr;
    state.x = st.x, state.y = st.y, state.z = st.z, state.step = 0;
    vis[st.x][st.y][st.z] = 1;
    q.push(state);
    state.x = ed.x, state.y = ed.y, state.z = ed.z, state.step = 0;
    visr[ed.x][ed.y][ed.z] = 1;
    qr.push(state);

    int STEP = 0, STEPR = 0;
    while (!q.empty() || !qr.empty())
    {
        while (!q.empty() && q.front().step <= STEP)
        {
            temp = q.front();
            q.pop();
            int x = temp.x, y = temp.y, z = temp.z, step = temp.step;
            int STEP = step;

            if (visr[x][y][z])
            {
                ans = STEP + STEPR;
                return;
            }

            for (int i = 0; i < 4; i++)
            {
                int xx = x + dx[i], yy = y + dy[i];
                if (0 <= xx && xx < r && 0 <= yy && yy < c && !vis[xx][yy][z] && G[z].g[xx][yy] != '#')
                {
                    vis[xx][yy][z] = 1;
                    state.x = xx, state.y = yy, state.z = z, state.step = step + 1;
                    q.push(state);
                }
            }
            if (z + 1 < l && !vis[x][y][z + 1] && G[z + 1].g[x][y] != '#')
            {
                vis[x][y][z + 1] = 1;
                state.x = x, state.y = y, state.z = z + 1;
                state.step = step + 1;
                q.push(state);
            }
            if (z - 1 >= 0 && !vis[x][y][z - 1] && G[z - 1].g[x][y] != '#')
            {
                vis[x][y][z - 1] = 1;
                state.x = x, state.y = y, state.z = z - 1;
                state.step = step + 1;
                q.push(state);
            }
        }
        if (!q.empty()) STEP++;
        /*---------------------------------------------------------------*/
        while (!qr.empty() && qr.front().step <= STEPR)
        {
            temp = qr.front();
            qr.pop();
            int x = temp.x, y = temp.y, z = temp.z, step = temp.step;

            if (vis[x][y][z])
            {
                ans = STEP + STEPR;
                return;
            }

            for (int i = 0; i < 4; i++)
            {
                int xx = x + dx[i], yy = y + dy[i];
                if (0 <= xx && xx < r && 0 <= yy && yy < c && !visr[xx][yy][z] && G[z].g[xx][yy] != '#')
                {
                    visr[xx][yy][z] = 1;
                    state.x = xx, state.y = yy, state.z = z, state.step = step + 1;
                    qr.push(state);
                }
            }
            if (z + 1 < l && !visr[x][y][z + 1] && G[z + 1].g[x][y] != '#')
            {
                visr[x][y][z + 1] = 1;
                state.x = x, state.y = y, state.z = z + 1;
                state.step = step + 1;
                qr.push(state);
            }
            if (z - 1 >= 0 && !visr[x][y][z - 1] && G[z - 1].g[x][y] != '#')
            {
                visr[x][y][z - 1] = 1;
                state.x = x, state.y = y, state.z = z - 1;
                state.step = step + 1;
                qr.push(state);
            }
        }
        if (!qr.empty()) STEPR++;
    }
}

int main()
{
    while (scanf("%d %d %d", &l, &r, &c) == 3 && (l + r + c))
    {
        for (int i = 0; i < l; i++)
        {
            getchar();
            for (int j = 0; j < r; j++)
            {
                for (int k = 0; k < c; k++)
                {
                    G[i].g[j][k] = getchar();
                    if (G[i].g[j][k] == 'S')
                    {
                        st.z = i, st.x = j, st.y = k;
                    }
                    else if (G[i].g[j][k] == 'E')
                    {
                        ed.z = i, ed.x = j, ed.y = k;
                    }
                }
                getchar();
            }
        }
        bfs();
        if (ans == -1)
        {
            printf("Trapped!\n");
        }
        else
        {
            printf("Escaped in %d minute(s).\n", ans);
        }
    }
    return 0;
}