POJ 1077-Eight(BFS+优先队列)

Eight

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 28463 Accepted: 12420 Special Judge

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 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

 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

题意：

  给出一个八数码给你，要你求出1，2，3，4，5，6，7，8，x的顺序。

思路：

  这题很明显用BFS来做，当我写好了代码才发现我竟然不会如何去剪枝，所以一直跑重复的点。之后去看了看题解发现直接用hash去保存九个位置就是这个算法的剪枝了，思维短路了，竟然没意识到应该用hash去剪枝。

AC代码：

#include<iostream>#include<functional>#include<algorithm>#include<cstring>#include<string>#include<vector>#include<cstdio>#include<queue>#include<cmath>#include<map>#include<set>using namespace std;#define CRL(a) memset(a,0,sizeof(a))#define QWQ ios::sync_with_stdio(0)#define inf 0x3f3f3f3ftypedef unsigned long long LL;typedef  long long ll;const int T = 500000+50;const int mod = 1000000007;char s[5][5];int xy[][2]={{1,0},{0,-1},{-1,0},{0,1}};char fx[] = {{'r'},{'u'},{'l'},{'d'}};struct node{char s[5][5];vector<char> ve;int c,x,y;node(){}bool operator<(const node& b)const{return c<b.c; }};set<int> se;inline int jugde(const node& b){int k=0;for(int i = 0;i<3;++i){for(int j=0;j<3;++j){if(b.s[i][j]=='0'+i*3+j+1)k++;}}return k;}inline void Copy1(node& a,const node& b){for(int i=0;i<3;++i){a.s[i][0]=b.s[i][0],a.s[i][1]=b.s[i][1],a.s[i][2]=b.s[i][2];a.s[i][3]='\0';}} int Hash(const node& a){int sum = 0;for(int i=0;i<3;++i){for(int j=0;j<3;++j){if(a.s[i][j]=='x'){sum = sum*10;}else {sum = sum*10 + a.s[i][j]-'0';}}}return sum;}void BFS(int x,int y){se.clear();priority_queue<node> q;node t,tt;for(int i=0;i<3;++i){t.s[i][0]=s[i][0],t.s[i][1]=s[i][1],t.s[i][2]=s[i][2];t.s[i][3]='\0';}t.x=x,t.y=y;t.c = jugde(t);q.push(t);se.insert(Hash(t));while(!q.empty()){t = q.top();q.pop();if(t.c==8){for(int i=0;i<t.ve.size();++i){printf("%c",t.ve[i]);}printf("\n");break;}for(int j=0;j<4;++j){int tx = t.x + xy[j][1];int ty = t.y + xy[j][0];if(tx>=0&&tx<3&&ty>=0&&ty<3){tt.x = tx;tt.y = ty;Copy1(tt,t);swap(tt.s[tx][ty],tt.s[t.x][t.y]);tt.c = jugde(tt);tt.ve = t.ve;tt.ve.push_back(fx[j]);int tmp = Hash(tt);if(se.find(tmp)==se.end()){se.insert(tmp);q.push(tt);}}}}}int main(){#ifdef zsc    freopen("input.txt","r",stdin);#endifint i,j,u,v;while(~scanf(" %c",&s[0][0])){for(i=1;i<9;++i){scanf(" %c",&s[i/3][i%3]);if(s[i/3][i%3]=='x')u=i/3,v=i%3;}BFS(u,v);}      return 0;}