Dungeon Master （广搜）

Dungeon Master

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 38   Accepted Submission(s) : 13

Problem 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). <br>L is the number of levels making up the dungeon. <br>R and C are the number of rows and columns making up the plan of each level. <br>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 <br><blockquote>Escaped in x minute(s). </blockquote><br>where x is replaced by the shortest time it takes to escape. <br>If it is not possible to escape, print the line <br><blockquote>Trapped! </blockquote>

 

Sample Input

3 4 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0

 

Sample Output

Escaped in 11 minute(s).Trapped!

 

Source

PKU

思路：

广搜题目，只不过由以前的4个方向改为6个方向！

代码：

#include <iostream>#include <stdio.h>#include <string.h>#include <queue>#include <algorithm>using namespace std;char map[35][35][35];int vis[35][35][35];int k,n,m,sx,sy,sz,ex,ey,ez;int dir[6][3] = {{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}};//六个方向struct node{    int x,y,z,step;};int check(int x,int y,int z)                //判断是否在界内以及是否出现过 {    if(x<0 || y<0 || z<0 || x>=k || y>=n || z>=m)        return 1;    else if(map[x][y][z] == '#')        return 1;    else if(vis[x][y][z])        return 1;    return 0;    }    int bfs()              //广搜    {        int i;        node a,next;        queue<node> Q;        a.x = sx;        a.y = sy;        a.z = sz;        a.step = 0;        vis[sx][sy][sz] = 1;        Q.push(a);        while(!Q.empty())        {            a = Q.front();            Q.pop();            if(a.x == ex && a.y == ey && a.z == ez)                return a.step;            for(i = 0; i<6; i++)            {                next = a;                next.x = a.x+dir[i][0];                next.y = a.y+dir[i][1];                next.z = a.z+dir[i][2];                if(check(next.x,next.y,next.z))                    continue;                vis[next.x][next.y][next.z] = 1;                next.step = a.step+1;                Q.push(next);            }        }        return 0;    }    int main()    {        int i,j,r;        while(cin>>k>>n>>m,n+m+k)        {            for(i = 0; i<k; i++)           //注意优化，其实完全可以不用三重循环            {                for(j = 0; j<n; j++)                {                   cin>>map[i][j];                    for(r = 0; r<m; r++)                    {                        if(map[i][j][r] == 'S')            //记录起点坐标                        {                            sx = i;                            sy = j;                            sz = r;                        }                        else if(map[i][j][r] == 'E')      //记录终点坐标                        {                            ex = i;                            ey = j;                            ez = r;                        }                    }                }            }            memset(vis,0,sizeof(vis));            int ans;            ans = bfs();            if(ans)                cout<<"Escaped in "<<ans<<" minute(s)."<<endl;            else                cout<<"Trapped!"<<endl;        }        return 0;    }

心得：

其实是很简单的广搜，但是wa了好几次，自己有思路，感觉程序没什么问题，但是就是不对！还得要细心！