hdu 2216 Game III

Game III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1285    Accepted Submission(s): 362

Problem Description

Zjt and Sara will take part in a game, named Game III. Zjt and Sara will be in a maze, and Zjt must find Sara. There are some strang rules in this maze. If Zjt move a step, Sara will move a step in opposite direction.
Now give you the map , you shold find out the minimum steps, Zjt have to move. We say Zjt meet Sara, if they are in the same position or they are adjacent . 
Zjt can only move to a empty position int four diraction (up, left, right, down). At the same time, Sara will move to a position in opposite direction, if there is empty. Otherwise , she will not move to any position.
The map is a N*M two-dimensional array. The position Zjt stays now is marked Z, and the position, where Sara stays, is marked E.

>  . : empty position
>  X: the wall
>  Z: the position Zjt now stay
>  S: the position Sara now stay

Your task is to find out the minimum steps they meet each other.

 

Input

The input contains several test cases. Each test case starts with a line contains three number N ,M (2<= N <= 20, 2 <= M <= 20 ) indicate the size of the map. Then N lines follows, each line contains M character. A Z and a S will be in the map as the discription above.

 

Output

For each test case, you should print the minimum steps. “Bad Luck!” will be print, if they can't meet each other.

 

Sample Input

4 4 XXXX .Z.. .XS. XXXX 4 4 XXXX .Z.. .X.S XXXX 4 4 XXXX .ZX. .XS. XXXX

 

Sample Output

1 1 Bad Luck!

 

Author

zjt

 

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個人表示這個題目還是很有意思的，其中比較費心思的是怎麼樣來標記每種狀態，後來看了別人的纔想通，應該用一個四維數組就可以了，這樣每個狀態就可以保證能夠是不是出現過了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<math.h>
using namespace std;
#define M 30
int maze[M][M][M][M];
char map[M][M];
int n,m,sx,sy,ex,ey;
struct node
{
    int ax,ay;
    int bx,by;
    int step;
}ua,ub;
int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{-1,0},{1,0},{0,-1},{0,1}};
int judge(node a)
{
     int x,y;
     x=fabs(a.bx-a.ax);
     y=fabs(a.by-a.ay);
     if(x+y<2)
     return 1;
     else
     return 0;
}
int check(int a,int b)
{
 if(a>=0&&a<n&&b>=0&&b<m&&map[a][b]!='X')
 return 1;
 return 0;
}
int bfs()
{
    queue<node>q;
    int x1,y1,x2,y2;
    int i,j;
    memset(maze,0,sizeof(maze));
    ua.ax=sx;
    ua.ay=sy;
    ua.bx=ex;
    ua.by=ey;
    ua.step=0;
    maze[sx][sy][ex][ey]=1;//以兩點的位置作爲標記，一旦走過就不再走了
    q.push(ua);
    while(!q.empty())
    {
        ua=q.front();
        q.pop();
        if(judge(ua))
        {
            return ua.step;
        }
        for(i=0;i<4;i++)
        {

            ub=ua;
            ub.ax=ub.ax+dir[i][0];
            ub.ay=ub.ay+dir[i][1];
            ub.step++;
            //z節點走到了某一個位置，下面判斷是否符合條件。
            if(check(ub.ax,ub.ay))//現在z所在的點符合了條件
            {
                if(check(ub.bx+dir[i+4][0],ub.by+dir[i+4][1]))
                {//s點進行了移動，進行計算標記
                    ub.bx=ub.bx+dir[i+4][0];
                    ub.by=ub.by+dir[i+4][1];
                    if(maze[ub.ax][ub.ay][ub.bx][ub.by]==0)
                    {
                        maze[ub.ax][ub.ay][ub.bx][ub.by]=1;
                        q.push(ub);
                    }





                }
                else
                {//s點沒有進行移動，但是目前的情況仍舊未曾發生過即可。

                     if(maze[ub.ax][ub.ay][ub.bx][ub.by]==0)
                    {
                        maze[ub.ax][ub.ay][ub.bx][ub.by]=1;
                        q.push(ub);
                    }





                }







            }







        }














    }






    return -1;


}
int main()
{
    int i,j,k;
    while(scanf("%d%d",&n,&m)!=EOF)
    {

        for(i=0;i<n;i++)
        {
          scanf("%s",&map[i]);
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(map[i][j]=='Z')
                {
                    sx=i;
                    sy=j;
                }
                else
                if(map[i][j]=='S')
                {
                    ex=i;
                    ey=j;
                }
            }
        }
        int ans=bfs();
        if(ans!=-1)
        {
           printf("%d\n",ans);
        }
        else
        {
            printf("Bad Luck!\n");
        }






    }
    return 0;
}