HDOJ題目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): 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代碼

#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_queueq;
	queueq;
	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