HDOJ question 3309 Roll The Cube (BFS)

HDOJ question 3309 Roll The Cube (BFS)
Roll The CubeTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 502 Accepted Submission (s): 181


Problem Description This is a simple game. The goal of the game is to roll two bils 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 bils at the same time. up, down, left, right --- once one of these keys is pressed, bils exist roll to that direction, for example, you pressed up, two bils both roll up.
A ball will stay where it is if its next point is a wall, and ballcan't be overlap.
Your code shoshould 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 bils (B) and two holes (H) in a map.
The boundary of the map is always Wils (*).
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

46 3****B**B**H**H****4 4*****BB**HH*****4 4*****BH**HB*****5 6*******.BB***.H*H**..*.*******

Sample Output

312Sorry , sir , my poor program fails to get an answer.

Author MadFroG
Source HDOJ Monthly Contest-2010.02.06
Recommend wxl | We have carefully selected several similar problems for you: 3308 3314 3307 3306 ac code

#include
 
  #include
  
   #include
   
    #include
    
     using namespace std;int n,m,vis[25][25][25][25],sx[2],sy[2];char map[25][25];int dx[4]={0,1,0,-1};int dy[4]={1,0,-1,0};struct s{int x[2],y[2],step,b[2],h[2];//friend bool operator <(s a,s b)//{//return a.step>b.step;//}}a,temp;int bfs(){memset(vis,0,sizeof(vis));a.x[0]=sx[0],a.x[1]=sx[1];a.y[0]=sy[0],a.y[1]=sy[1];a.b[0]=a.b[1]=a.h[0]=a.h[1]=0;vis[sx[0]][sy[0]][sx[1]][sy[1]]=1;a.step=0;//priority_queue
     
      q;queue
      
       q;q.push(a);while(!q.empty()){int i,j;//a=q.top();a=q.front();q.pop();for(i=0;i<4;i++){temp=a;for(j=0;j<2;j++){if(temp.b[j])continue;temp.x[j]=a.x[j]+dx[i];temp.y[j]=a.y[j]+dy[i];if(map[temp.x[j]][temp.y[j]]=='*'){temp.x[j]=a.x[j];temp.y[j]=a.y[j];}}if(vis[temp.x[0]][temp.y[0]][temp.x[1]][temp.y[1]])continue;if(temp.x[0]==temp.x[1]&&temp.y[0]==temp.y[1]&&temp.b[0]+temp.b[1]==0)continue;vis[temp.x[0]][temp.y[0]][temp.x[1]][temp.y[1]]=1;temp.step=a.step+1;int flag=1;for(j=0;j<2;j++){int now=map[temp.x[j]][temp.y[j]];if(now<2&&!temp.h[now]){temp.h[now]=1;temp.b[j]=1;}if(!temp.b[j])flag=0;}if(flag)return temp.step;q.push(temp);}}return -1;}int main(){int t;scanf("%d",&t);while(t--){int i,j;scanf("%d%d",&n,&m);int cnt=0,cot=0;for(i=0;i