Hdu 3309 Roll The Cube

Roll The Cube

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 715    Accepted Submission(s): 295

Problem Description

This is a simple game.The goal of the game is to roll two balls to two holes each.
'B' -- ball
'H' -- hole
'.' -- land
'*' -- wall
Remember when a ball rolls into a hole, they(the ball and the hole) disappeared, that is , 'H' + 'B' = '.'.
Now you are controlling two balls at the same time.Up, down , left , right --- once one of these keys is pressed, balls exist roll to that direction, for example , you pressed up , two balls both roll up.
A ball will stay where it is if its next point is a wall, and balls can't be overlap.
Your code should give the minimun times you press the keys to achieve the goal.

 

 

Input

First there's an integer T(T<=100) indicating the case number.
Then T blocks , each block has two integers n , m (n , m <= 22) indicating size of the map.
Then n lines each with m characters.
There'll always be two balls(B) and two holes(H) in a map.
The boundary of the map is always walls(*).

 

 

Output

The minimum times you press to achieve the goal.
Tell me "Sorry , sir , my poor program fails to get an answer." if you can never achieve the goal.

 

 

Sample Input

4

6 3

***

*B*

*B*

*H*

*H*

***

 

4 4

****

*BB*

*HH*

****

 

4 4

****

*BH*

*HB*

****

 

5 6

******

*.BB**

*.H*H*

*..*.*

******

 

 

Sample Output

3

1

2

Sorry , sir , my poor program fails to get an answer.

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;

struct node
{
    int x[2],y[2],step;
    int b[2],h[2];
};
char map[25][25];
bool vis[25][25][25][25];
int sx[2],sy[2],n,m;
int to[4][2] = {0,1,1,0,0,-1,-1,0};

int bfs()
{
    queue<node> Q;
    node a,next;
    a.x[0] = sx[0],a.y[0]=sy[0];//两个球     a.x[1] = sx[1],a.y[1]=sy[1];
    a.step = 0;
    a.b[0] = a.b[1] = a.h[0] = a.h[1] = 0;
    Q.push(a);
    memset(vis,false,sizeof(vis));
    while(!Q.empty())
    {
        a = Q.front();
        Q.pop();
        int i,j;
        for(i = 0; i<4; i++)
        {
            next = a;
            for(j = 0; j<2; j++)
            {
                if(next.b[j]) continue;//如果球已经进洞了，标记一下                 next.x[j]+=to[i][0];//网其他方向搜索                 next.y[j]+=to[i][1];
                if(map[next.x[j]][next.y[j]]=='*')//如果是墙，保持不动，也就是坐标会退到上一个方向                 {
                    next.x[j]-=to[i][0];
                    next.y[j]-=to[i][1];
                }
            }
            if(vis[next.x[0]][next.y[0]][next.x[1]][next.y[1]])//已经访问过                 continue;
            if(next.x[0]==next.x[1]&& next.y[0]==next.y[1] && next.b[0]+next.b[1]==0)//两个球相撞                 continue;
            next.step++;
            vis[next.x[0]][next.y[0]][next.x[1]][next.y[1]] = true;//标记为走过             int flag = 1;
            for(j = 0; j<2; j++)
            {
                int now = map[next.x[j]][next.y[j]];
                if(now<2 && !next.h[now])//如果球到了洞的位置                 {
                    next.h[now]=1;//球进洞                     next.b[j] = 1;
                }
                if(!next.b[j])//如果洞找到完了                     flag = 0;
            }
            if(flag)
                return next.step;
            Q.push(next);
        }
    }
    return -1;
}

int main()
{
    int t,i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        int cnt = 0,len = 0;
        for(i = 0; i<n; i++)
        {
            scanf("%s",map[i]);
            for(j = 0; j<m; j++)
            {
                if(map[i][j]=='H') map[i][j] = cnt++;
                else if(map[i][j]=='B')//标记球的位置                 {
                    sx[len] = i,sy[len] = j;
                    len++;
                }
            }
        }
        int ans = bfs();
        if(ans!=-1)
            printf("%d\n",ans);
        else
            printf("Sorry , sir , my poor program fails to get an answer.\n");
    }

    return 0;
}