HDU 3309 Roll The Cube题解

Roll The Cube

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

Total Submission(s): 273    Accepted Submission(s): 110

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.
 
 
本题是一般的BFS，只要记着当两个球重合时，只要一个球在洞里，那么另一个求解可以从这里经过，还有就是两球是同时运动的
View Code 
  1 #include<iostream>
  2 using namespace std;
  3 #include<queue>
  4 #include<cstdio> 
  5 int dir[4][2]={1,0,0,-1,-1,0,0,1}; 
  6 struct node
  7 {
  8     int x[2],y[2];
  9     int step;
 10     node(){}; 
 11     node(int xx1,int yy1,int xx2,int yy2,int vv){x[0]=xx1;y[0]=yy1;x[1]=xx2;y[1]=yy2;step=vv;}    
 12 };
 13 
 14 node st;
 15 int n,m,cnt=0;
 16 
 17 
 18 int use[22][22][22][22]={0};
 19 char map[22][22];
 20 inline bool check(int x,int y)
 21 {
 22 if(x>=0&&x<n&&y>=0&&y<m)return 1;
 23 return 0; 
 24 } 
 25 
 26 
 27 inline bool equal(int x1,int y1,int x2,int y2)
 28 {
 29       return (x1==x2&&y1==y2); 
 30 } 
 31 
 32 int BFS()
 33 {
 34     queue<node> q;
 35     q.push(st);    
 36     node cur;
 37     while(!q.empty())
 38     {
 39         cur=q.front();
 40         q.pop();
 41         int x1,y1,x2,y2;
 42         int flag=0;
 43         for(int i=0;i<4;i++)
 44         {
 45             if(map[cur.x[0]][cur.y[0]]=='H')
 46             {
 47                 x1=cur.x[0];
 48                 y1=cur.y[0];
 49                 flag=1;    
 50             }
 51             else
 52             {
 53                 x1=cur.x[0]+dir[i][0];
 54                 y1=cur.y[0]+dir[i][1];    
 55             }
 56             if(check(x1,y1)==0||map[x1][y1]=='*')
 57             {
 58                 x1=cur.x[0];
 59                 y1=cur.y[0];
 60             }
 61             
 62             if(map[cur.x[1]][cur.y[1]]=='H'&&!flag)
 63             {
 64                 x2=cur.x[1];
 65                 y2=cur.y[1];    
 66             }
 67             else
 68             {
 69                 x2=cur.x[1]+dir[i][0];
 70                 y2=cur.y[1]+dir[i][1];    
 71             }
 72             
 73             
 74             if(check(x2,y2)==0||map[x2][y2]=='*')
 75             {
 76                 x2=cur.x[1];
 77                 y2=cur.y[1];
 78             }
 79             
 80             if(use[x1][y1][x2][y2]==cnt) continue;
 81             if(equal(x1,y1,x2,y2))
 82             {
 83                     if(map[x1][y1]=='H'||map[x2][y2]=='H')
 84                     {
 85                             use[x1][y1][x2][y2]=cnt;
 86                             use[x2][y2][x1][y1]=cnt;
 87                             q.push(node(x1,y1,x2,y2,cur.step+1));     
 88                     }    
 89             }
 90             else
 91             {
 92                        if(map[x1][y1]=='H'&&map[x2][y2]=='H')return cur.step+1;
 93                        use[x1][y1][x2][y2]=cnt;
 94                        use[x2][y2][x1][y1]=cnt;
 95                       q.push(node(x1,y1,x2,y2,cur.step+1));     
 96             }        
 97         }    
 98     }
 99     return -1; 
100 } 
101 
102 int main()
103 {
104     int kase,k;
105     scanf("%d",&kase);
106     while(kase--)
107     {
108             scanf("%d%d",&n,&m);
109             cnt++; 
110             for(int i=0;i<n;i++)
111             scanf("%s",map[i]);
112             k=0; 
113             for(int i=0;i<n&&k<2;i++)
114             for(int j=0;j<m&&k<2;j++)
115             {
116                     if(map[i][j]=='B')
117                     {
118                         st.x[k]=i;
119                         st.y[k]=j;
120                         map[i][j]='.';
121                         k++;
122                         if(k==2)
123                         break;    
124                     }    
125             }
126             
127             st.step=0;
128             use[st.x[0]][st.y[0]][st.x[1]][st.y[1]]=cnt;
129             use[st.x[1]][st.y[1]][st.x[0]][st.y[0]]=cnt;
130         int    ans=BFS(); 
131         if(ans==-1)printf("Sorry , sir , my poor program fails to get an answer.\n");
132         else printf("%d\n",ans);     
133     } 
134     return 0;    
135 }

 

posted on 2012-08-12 12:14  我的ACM之路  阅读(440)  评论(0)    收藏  举报

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