Eight(单向搜索)
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http://poj.org/problem?id=1077
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:
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:
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.
1 2 3 4 5 6 7 8 9 10 11 12 13 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 12 13 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 a description of a configuration of the 8 puzzle. The 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
is described by this list:
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.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
Source
South Central USA 1998
用各种方法都写一下。
又写了一个新的代码:
/*POJ 1077 Eight单向搜索,从正向开始找目标点康托展开作为hash值AC G++ 8876K 96MS*/#include<stdio.h>#include<string.h>#include<iostream>#include<string>using namespace std;const int MAXN=362881;//9!=362880struct Node{ int s[9]; int pre;//记录前一个结点 int dir;//记录前一个结点到该结点的方向}que[MAXN];bool hash[MAXN];int path[MAXN];int fac[]={1,1,2,6,24,120,720,5040,40320,362880};//康拖展开判重// 0!1!2!3! 4! 5! 6! 7! 8! 9!int cantor(int *s){ int sum=0; for(int i=0;i<9;i++) { int num=0; for(int j=i+1;j<9;j++) if(s[j]<s[i]) num++; sum+=(num*fac[9-i-1]); } return sum;}int move[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//u,d,l,rvoid output(int len){ for(int i=len-1;i>=0;i--) { if(path[i]==0) printf("u"); else if(path[i]==1) printf("d"); else if(path[i]==2) printf("l"); else if(path[i]==3) printf("r"); } printf("\n");}void bfs(){ int front=-1,rear=0; que[0].pre=-1; que[0].dir=-1; memset(hash,false,sizeof(hash)); hash[cantor(que[0].s)]=true; if(cantor(que[0].s)==0){output(0);return;} while(front<rear) { front++; int tmp; for(tmp=0;tmp<9;tmp++) if(que[front].s[tmp]==9) break; int x=tmp/3; int y=tmp%3; for(int i=0;i<4;i++) { int tx=x+move[i][0]; int ty=y+move[i][1]; if(tx<0||tx>2||ty<0||ty>2)continue; que[rear+1]=que[front]; que[rear+1].pre=front; que[rear+1].dir=i; que[rear+1].s[tmp]=que[rear+1].s[tx*3+ty]; que[rear+1].s[tx*3+ty]=9; int now=cantor(que[rear+1].s); if(now==0)//到达目标 { int len=0; int t=rear+1; while(que[t].pre!=-1) { path[len++]=que[t].dir; t=que[t].pre; } output(len); return; } if(!hash[now]) { rear++; hash[now]=true; } } }}int main(){ // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); char str[10]; while(scanf("%s",&str)!=EOF) { if(str[0]=='x') que[0].s[0]=9; else que[0].s[0]=str[0]-'0'; for(int i=1;i<9;i++) { scanf("%s",&str); if(str[0]=='x') que[0].s[i]=9; else que[0].s[i]=str[0]-'0'; } bfs(); } return 0;}
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