Roll The Cube

Time 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 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
Recommend wxl | We have carefully selected several similar problems for you: 3308 3314 3307 3306 3310 ac代码

#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

HDOJ题目3309 Roll The Cube（BFS）

Roll The Cube