hdu 1043 A*算法解决八数码问题

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23880    Accepted Submission(s): 6384
Special Judge

Problem Description

The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:

 1  2  3  4 5  6  7  8 9 10 11 1213 14 15  x

where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:

 1  2  3  4     1  2  3  4     1  2  3  4     1  2  3  4 5  6  7  8     5  6  7  8     5  6  7  8     5  6  7  8 9  x 10 12     9 10  x 12     9 10 11 12     9 10 11 1213 14 11 15    13 14 11 15    13 14  x 15    13 14 15  x            r->            d->            r->

The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.

Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).

In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.

 

Input

You will receive, several descriptions of configuration of the 8 puzzle. One description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle

1 2 3
x 4 6
7 5 8

is described by this list:

1 2 3 x 4 6 7 5 8

 

Output

You will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line. Do not print a blank line between cases.

 

Sample Input

2  3  4  1  5  x  7  6  8

 

Sample Output

ullddrurdllurdruldr

 

Source

South Central USA 1998 (Sepcial Judge Module By JGShining)

 

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题意：给出3*3的方格，每个方格中有一个字母，且各不相同，来自于数字[0:8]或字母x，给出最快的操作序列使得排布变为[0:8],x

解法：A*算法,f=g+h,h为当前状态到目标状态的一个估计值，要<=实际花费值，g为当前实际花费步数，用优先队列对状态结点进行排序+用康托展开保存状态。

剪枝：在进行搜索之前，对除X之外的数列求逆序数，如果为奇数，那么不可能到达目标状态，因为目标状态的逆序值必定是偶数，而且每移动一步，当前状态除X之外的数列的逆序数不变。

原因是：

1.如果X和左边的数交换，那么数列本身不变(因为没有统计X)，故逆序数自然不变。

2.如果X和上边的数交换：相当于把  ... a b c  ... 变成了...b c a ... ，那么必然可以忽略掉左右的省略号部分。

假如a比b、c都大，记f(x)为x左边比x大的个数。那么移动后f(a)不变，f(b)-1,f(c)-1,故逆序数奇偶性不变。

假如a比b、c都小，那么f(a)+2，逆序数奇偶性不变。

else，即假如a比其中一个大，比其中一个小，那么f(a)+1，还有一个f()-1，另外一个f()不变，故逆序数的奇偶性不变。

#include<iostream>#include<cstdio>#include<cstring>#include<queue>#include<algorithm>using namespace std;char s[50];int b[9];int ha[9]={1,1,2,6,24,120,720,5040,40320};  char move_name[5]="udlr";int dir[4][2]={{-1,0},{+1,0},{0,-1},{0,+1}};struct Node{int h,g,x,y,hash_num;int a[9];bool operator<(const Node c)const{return h+g>=c.h+c.g;}  int get_inverse(){int cnt=0;for(int i=0;i<9;i++)  if(a[i]){for(int j=0;j<i;j++) if(a[j]>a[i]) cnt++;}return cnt++;}int get_hash(){hash_num=0;for(int i=0;i<9;i++)  {int cnt=0;for(int j=0;j<i;j++)  if(a[j]>a[i])  cnt++;hash_num+=ha[i]*cnt; } return hash_num;}int get_h(){h=0;for(int i=0;i<9;i++) if(a[i]){int px=(a[i]-1)/3,py=(a[i]-1)%3;h+=abs(px-i/3)+abs(py-i%3);}return h;}bool eq(Node c){return hash_num==c.hash_num;}}S,E;const int maxn=4e5+4;struct HSET{int pre;char mov;} hset[maxn];bool vis[maxn];bool in(int x,int y){return 0<=x&&x<=2&&0<=y&&y<=2;}int get_pos(int x,int y){return 3*x+y;}void print_ans(int ind){if(hset[ind].pre==-1) return;print_ans(hset[ind].pre);putchar(hset[ind].mov);}void Astar(){if(S.eq(E)) return;priority_queue<Node>q;q.push(S);while(!q.empty()){Node nod=q.top();q.pop();int x=nod.x,y=nod.y;int pos=get_pos(x,y);for(int i=0;i<4;i++){int tx=x+dir[i][0],ty=y+dir[i][1];if(!in(tx,ty))  continue;Node nex=nod;int tpos=get_pos(tx,ty);swap(nex.a[pos],nex.a[tpos]);int hsh=nex.get_hash();// printf("%d %d",hsh,nod.hash_num);if(vis[hsh])  continue;vis[hsh]=1;hset[hsh].pre=nod.hash_num;hset[hsh].mov=move_name[i];nex.x=tx,nex.y=ty;nex.g++;nex.get_h();if(nex.eq(E)){print_ans(hsh);return;}q.push(nex);}}}int main(){for(int i=0;i<8;i++) E.a[i]=i+1;E.a[8]=0;E.x=E.y=2;E.get_hash();while(gets(s)){memset(vis,0,sizeof vis);int len=strlen(s);int k=0;for(int i=0;i<len;i++){if(s[i]==' ')  continue;if(s[i]=='x')  {S.a[k]=0; S.x=k/3; S.y=k%3; k++; }else    S.a[k++]=s[i]-'0';}S.g=0;S.get_h();// for(int i=0;i<9;i++)  printf("%d ",S.a[i]);int ivs=S.get_inverse();if(ivs&1){puts("unsolvable");continue;}int hsh=S.get_hash();hset[hsh].pre=-1;vis[hsh]=1;Astar();putchar('\n');}return 0;}