uva 10085 - The most distant state

Problem A

The Most Distant State

Input: standard input

Output: standard output

 

The 8-puzzle is a square tray in which eight square tiles are placed. The remaining ninth square is uncovered. Each tile has a number on it. A tile that is adjacent to the blank space can be slid into that space. A game consists of a starting state and a specified goal state. The starting state can be transformed into the goal state by sliding (moving) the tiles around. The 8-puzzle problem asks you to do the transformation in minimum number of moves.

 

2	8	3				1	2	3
1	6	4	=>			8		4
7		5				7	6	5
Start						Goal

 

However, our current problem is a bit different. In this problem, given an initial state of the puzzle your are asked to discover a goal state which is the most distant (in terms of number of moves) of all the states reachable from the given state.

Input

The first line of the input file contains an integer representing the number of test cases to follow. A blank line follows this line.

Each test case consists of 3 lines of 3 integers each representing the initial state of the puzzle. The blank space is represented by a 0 (zero). A blank line follows each test case.

 

Output

For each test case first output the puzzle number. The next 3 lines will contain 3 integers each representing one of the most distant states reachable from the given state. The next line will contain the shortest sequence of moves that will transform the given state to that state. The move is actually the movement of the blank space represented by four directions : U (Up), L (Left), D (Down) and R (Right). After each test case output an empty line.

 

Sample Input

1

2 6 4
1 3 7
0 5 8

Sample Output

Puzzle #1
8 1 5
7 3 6
4 0 2
UURDDRULLURRDLLDRRULULDDRUULDDR

要求從起始狀態（已知）到目標狀態（未知）移動次數儘可能多，當然不能重複移動所以要判重，沒想到高效的算法，只能把所以可以移動的都求出來，那麼最後一次不能移動自然就是目標狀態，他的移動次數一定是最多的，bfs出所有的狀態；判重用全排列的哈希，狀態數目太多了bfs寫成遞歸調用貌似會棧溢出的樣子；

#include<string.h>
#include<stdio.h>
#define P   for (i=0;i<9;i++){printf("%d",q[top].a[i]);if ((i+1)%3==0) printf("\n");else   printf(" ");}
struct node
{int a[9],x,y,succ,dir;
}q[362881];
int top,tail,visit[362881],exp[8]={40320,5040,720,120,24,6,2,1},
     move[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
int s[1000];
int hash(struct node *A,int num )//全排列哈希
{int i,j,x,y=0;
 for (i=0;i<9;i++)
 {
  x=0;
  for (j=i+1;j<9;j++)
  if (q[num].a[i]>q[num].a[j]) ++x; //統計逆序數
  y=y+x*exp[i];
 }
 return y;
}

int main()
{begin
 int i,j,k,t,T,px,py,pk,sum,temp;
 scanf("%d",&T);
 for (t=1;t<=T;t++)
 {
  memset(visit,0,sizeof(visit));
  for (i=0;i<9;i++)
  {
   scanf("%d",&q[0].a[i]);
   if (q[0].a[i]==0) {q[0].x=i/3;q[0].y=i%3;}
  }

  top=0; tail=0;
  visit[hash(&q[0],0)]=1;

  while (top<=tail)
  {
   for (i=0;i<4;i++)
   {
    px=q[top].x+move[i][0];
    py=q[top].y+move[i][1];
    if ((px>=0)&&(py>=0)&&(px<3)&&(py<3))
    {
     k=q[top].x*3+q[top].y;
     pk=px*3+py;
     ++tail;
     for (j=0;j<9;j++)
     q[tail].a[j]=q[top].a[j];
     temp=q[tail].a[k]; q[tail].a[k]=q[tail].a[pk]; q[tail].a[pk]=temp;
     k=hash(&q[tail],tail);
     if (visit[k]) --tail;
     else {visit[k]=1;q[tail].succ=top; q[tail].dir=i;q[tail].x=px;q[tail].y=py;}
    }
   }
   ++top;
  }
  --top;
  printf("Puzzle #%d\n",t);P;
  sum=0;
  while (top)
  {
   s[++sum]=q[top].dir;
   top=q[top].succ;
  }
  for (i=sum;i>=1;i--)
  {
   if (s[i]==0) printf("U");
   if (s[i]==1) printf("D");
   if (s[i]==2) printf("L");
   if (s[i]==3) printf("R");
  }
  printf("\n\n");
 }
 end;
 return 0;
}