ZOJ 1940 Dungeon Master （三维广搜）

Dungeon Master

Time Limit: 2 Seconds Memory Limit: 65536 KB

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!

比赛时一看像搜索题，我就直接把这道题跳过去做其他题了。比赛完听队友说其实这道题很简单，就是一个简单地广搜。但是因为之前我没怎么学过搜索，所以并不后悔。只是我该好好学习搜索了。

题意：一个三维的地牢，给你起点和终点，问从起点到终点至少需要多长时间，每移动一次耗时1minute；如果不能到达终点，输出“Trapped！“。因为是三维的，所以从 每个点开始搜索时都会有6个方向，设置6个方向向量，然后每搜到一个点就判断是不是终点。利用队列即可。
下面是我的AC代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<iostream>
#include<algorithm>
using namespace std;

int dx[6] = {0,0,-1,1,0,0};
int dy[6] = {0,0,0,0,1,-1};
int dz[6] = {1,-1,0,0,0,0};

char Map[40][40][40];
int vis[40][40][40], L, R, C;

struct node
{
    int x, y, z;
    int time;
}st, ed;
queue<node> q;

bool check(int x, int y, int z) 
{
    if(x >= 1 && x <= L && y >= 1 && y <= R && z >= 1 && z <= C)
        return true;
    return false;
}

int BFS()
{
    int x, y, z, t, i;
    while(!q.empty())
    {
        node tmp = q.front();
        q.pop();
        x = tmp.x;
        y = tmp.y;
        z = tmp.z;
        t = tmp.time;
        for(i = 0; i < 6; i++)
        {
            int nx = x + dx[i];
            int ny = y + dy[i];
            int nz = z + dz[i];
            if(!vis[nx][ny][nz] && Map[nx][ny][nz] != '#' && check(nx,ny,nz))
            {
                if(nx == ed.x && ny == ed.y && nz == ed.z)
                    return t+1;
                vis[nx][ny][nz] = 1;
                node temp;
                temp.x = nx;
                temp.y = ny;
                temp.z = nz;
                temp.time = t + 1;
                q.push(temp);
            }
        }
    }
    return -1;
}

int main()
{
    while(~scanf("%d%d%d",&L, &R, &C) && (L + R + C))
    {
        memset(vis,0,sizeof(vis));
        int i, j, k;
        for(i = 1; i <= L; i++)
        {
            for(j = 1; j <= R; j++)
            {
                for(k = 1; k <= C; k++)
                {

                    cin>>Map[i][j][k];
                    if(Map[i][j][k] == 'S')
                    {
                        st.x = i, st.y = j, st.z = k;st.time = 0;
                        q.push(st);
                         vis[i][j][k] = 1;
                    }
                    else if(Map[i][j][k] == 'E')
                        ed.x = i, ed.y = j, ed.z = k;
                }
            }
        }
        int ans = BFS();
        if(ans == -1)
            printf("Trapped!\n");
        else
            printf("Escaped in %d minute(s).\n",ans);
        while(!q.empty())  q.pop();
    }
    return 0;
}

后来队友和我说了他的做法：不必判断搜到的点是否越界，只需先预处理一下，把可以经过的点设为1，不能经过的点即是墙的点设为0，然后存地图的时候从第1行，第1列，第1层开始存，这样边界都是0，即都是墙。搜索时就不用判断是否越界了。

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<iostream>
using namespace std;

int dx[6] = {0,0,0,0,1,-1};
int dy[6] = {0,0,1,-1,0,0};
int dz[6] = {1,-1,0,0,0,0};

int vis[40][40][40], Map[40][40][40];
struct node
{
    int x, y, z, step;
}st, ed;
queue<node> Q;

bool judge(int x, int y, int z) //判断当前位置是否可以访问
{
    if(!Map[x][y][z] || vis[x][y][z])
        return false;
    return true;
}

int BFS()
{
    int x, y, z, t, i;
    while(!Q.empty())
    {
        node tmp = Q.front();
        Q.pop();
        x = tmp.x;
        y = tmp.y;
        z = tmp.z;
        t = tmp.step;
        for(i = 0; i < 6; i++)
        {
            int nx = x + dx[i];
            int ny = y + dy[i];
            int nz = z + dz[i];
            if(judge(nx,ny,nz))
            {
                 if(nx == ed.x && ny == ed.y && nz == ed.z)
                    return t + 1;
                 vis[nx][ny][nz] = 1;
                 node temp;
                 temp.x = nx;
                 temp.y = ny;
                 temp.z = nz;
                 temp.step = t + 1;
                 Q.push(temp);
            }
        }
    }
    return -1;
}

int main()
{
    int L, R, C, i, j, k;
    char ch;
    while(~scanf("%d%d%d",&L, &R, &C) && (L + R + C))
    {
        while(!Q.empty())  Q.pop();
        memset(vis,0,sizeof(vis));
        memset(Map,0,sizeof(Map));
        for(i = 1; i <= L; i++)
            for(j = 1; j <= R; j++)
                for(k = 1; k <= C; k++)
                {
                    cin >> ch;
                    if(ch == '.')
                        Map[j][k][i] = 1;
                    else if(ch == 'S')
                    {
                        st.x = j;
                        st.y = k;
                        st.z = i;
                        st.step = 0;
                        vis[j][k][i] = 1;
                        Map[j][k][i] = 1;
                        Q.push(st);
                    }
                    else if(ch == 'E')
                    {
                        ed.x = j;
                        ed.y = k;
                        ed.z = i;
                        Map[j][k][i] = 1;
                    }
                }
        int ans = BFS();
        if(ans == -1)
            printf("Trapped!\n");
        else
            printf("Escaped in %d minute(s).\n",ans);
    }
    return 0;
}