hdu 3309 Roll The Cube ( bfs )

Roll The Cube

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 345    Accepted Submission(s): 127

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.
   
   
   
   

 

Author

MadFroG

 

Source

HDOJ Monthly Contest – 2010.02.06

思路：

bfs，用两个球的位置来判重，其他的直接模拟就ok了。最好将球和洞分开看，还有就是注意一下两个球不能在同一个位置。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define maxn 25
#define INF 0x3f3f3f3f
using namespace std;

int n,m,ans;
int sx[2],sy[2];
int dx[]= {-1,1,0,0};
int dy[]= {0,0,-1,1};
bool vis[maxn][maxn][maxn][maxn];
char mp[maxn][maxn];
char s[maxn];
struct Node
{
    int x[2],y[2],step;
    int b[2],h[2];   // 球 洞  球是否进洞和洞是否被求填满分开看
} cur,now;
queue<Node>q;

bool bfs()
{
    int i,j,t,flag;
    memset(vis,0,sizeof(vis));
    while(!q.empty()) q.pop();
    cur.x[0]=sx[0],cur.y[0]=sy[0];
    cur.x[1]=sx[1],cur.y[1]=sy[1];
    cur.h[0]=cur.h[1]=0;
    cur.b[0]=cur.b[1]=0;
    cur.step=0;
    vis[sx[0]][s[0]][sx[1]][sy[1]]=1;
    q.push(cur);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        for(i=0; i<4; i++)
        {
            cur=now;
            for(j=0; j<2; j++)
            {
                if(cur.b[j]) continue ;
                cur.x[j]+=dx[i];
                cur.y[j]+=dy[i];
                if(mp[cur.x[j]][cur.y[j]]=='*')
                {
                    cur.x[j]-=dx[i];
                    cur.y[j]-=dy[i];
                }
            }
            if(vis[cur.x[0]][cur.y[0]][cur.x[1]][cur.y[1]]||cur.x[0]==cur.x[1]&&cur.y[0]==cur.y[1]&&cur.b[0]+cur.b[1]==0) continue ;
            cur.step++;
            vis[cur.x[0]][cur.y[0]][cur.x[1]][cur.y[1]]=1;
            flag=1;
            for(j=0; j<2; j++)
            {
                t=mp[cur.x[j]][cur.y[j]];
                if(t<2&&!cur.h[t]) cur.b[j]=1,cur.h[t]=1;
                if(!cur.b[j]) flag=0;
            }
            if(flag)
            {
                ans=cur.step;
                return true ;
            }
            q.push(cur);
        }
    }
    return false ;
}
int main()
{
    int i,j,t,cnt1,cnt2;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        cnt1=cnt2=0;
        for(i=1; i<=n; i++)
        {
            scanf("%s",s);
            for(j=1; j<=m; j++)
            {
                mp[i][j]=s[j-1];
                if(mp[i][j]=='H') mp[i][j]=cnt1++;
                else if(mp[i][j]=='B') sx[cnt2]=i,sy[cnt2]=j,cnt2++;
            }
        }
        if(bfs()) printf("%d\n",ans);
        else printf("Sorry , sir , my poor program fails to get an answer.\n");
    }
    return 0;
}