八数码

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

经典的八数码问题,不过这道题的难点主要在于时间限制,如果时间开放为十秒,估计问题也就没那么复杂了,直接用stl中的map去重及标记。

可是时间限制为一秒,这就说明无法采用stl,只能用hash+BFS

开始的时候用hash+BFS 做了一个:

#include"iostream" #include"cstring" #include"queue" #include"stack" using namespace std; const int maxn=370000; int book[maxn]; struct node { int status[9]; //记录当前状态 int log; //记录x的位置 int hashd; //记录当前状态哈希值 string path; //记录路径 }; int goal[]={1,2,3,4,5,6,7,8,0}; int aim; int mov1[4][2]={{-1,0},{1,0},{0,-1},{0,1}}; string mov2="udlr"; string ans; int HASH[10]={1,1,2,6,24,120,720,5040,40320,362880}; int cantor(int a[]) { int answer=0; int number=0; for(int i=0;i<9;i++) { number=0; for(int j=i+1;j<9;j++) if(a[i]>a[j]) number++; answer+=number*HASH[8-i]; } return answer+1; } int BFS(struct node c) { queue<struct node> que; que.push(c); book[c.hashd]=1; while(!que.empty()) { struct node cur,nex; cur=que.front(); if(cur.hashd==aim) { ans=cur.path; // for(int kk=0;kk<9;kk++) cout<<cur.status[kk]<<' '; return 1; } que.pop(); int x=cur.log/3; int y=cur.log%3; for(int k=0;k<4;k++) { int tx=x+mov1[k][0]; int ty=y+mov1[k][1]; if(tx<0||tx>2||ty<0||ty>2) continue; nex=cur; nex.log=tx*3+ty; nex.status[cur.log]=nex.status[nex.log]; nex.status[nex.log]=0; nex.hashd=cantor(nex.status); if(book[nex.hashd]==0) { book[nex.hashd]=1; nex.path+=mov2[k]; que.push(nex); } } } return 0; } int main() { struct node ncur; char ch; aim=cantor(goal); // cout<<aim<<endl; cin>>ch; memset(book,0,sizeof(book)); if(ch=='x') { ncur.status[0]=0; ncur.log=0; } else { ncur.status[0]=ch-'0'; } for(int i=1;i<=8;i++) { cin>>ch; if(ch=='x') { ncur.status[i]=0; ncur.log=i; } else { ncur.status[i]=ch-'0'; } } ncur.hashd=cantor(ncur.status); if(BFS(ncur)) cout<<ans<<endl; else cout<<"unsolvable"<<endl; }
结果超时,后来将每次都记录字符串改为记录单个的字符,过了

#include"iostream" #include"cstring" #include"queue" #include"cstdio" #include"stack" using namespace std; const int maxn=370000; struct note { int flag; int f; char path; }; struct note book[maxn]; struct node { int status[9]; //记录当前状态 int log; //记录x的位置 int hashd; //记录当前状态哈希值 }; int goal[]={1,2,3,4,5,6,7,8,0}; int aim; int mov1[4][2]={{-1,0},{1,0},{0,-1},{0,1}}; string mov2="udlr"; string ans; int HASH[10]={1,1,2,6,24,120,720,5040,40320,362880}; int cantor(int a[]) { int answer=0; int number=0; for(int i=0;i<9;i++) { number=0; for(int j=i+1;j<9;j++) if(a[i]>a[j]) number++; answer+=number*HASH[8-i]; } return answer+1; } int BFS(struct node c) { queue<struct node> que; que.push(c); book[c.hashd].f=0; book[c.hashd].flag=1; book[c.hashd].path='k'; while(!que.empty()) { struct node cur,nex; cur=que.front(); if(cur.hashd==aim) { // ans=cur.path; // for(int kk=0;kk<9;kk++) cout<<cur.status[kk]<<' '; return 1; } que.pop(); int x=cur.log/3; int y=cur.log%3; for(int k=0;k<4;k++) { int tx=x+mov1[k][0]; int ty=y+mov1[k][1]; if(tx<0||tx>2||ty<0||ty>2) continue; nex=cur; nex.log=tx*3+ty; nex.status[cur.log]=nex.status[nex.log]; nex.status[nex.log]=0; nex.hashd=cantor(nex.status); if(book[nex.hashd].flag==0) { book[nex.hashd].flag=1; book[nex.hashd].f=cur.hashd; // nex.path+=mov2[k]; book[nex.hashd].path=mov2[k]; que.push(nex); } } } return 0; } int main() { char ch; aim=cantor(goal); // cout<<aim<<endl; while(cin>>ch) { struct node ncur; memset(book,0,sizeof(book)); if(ch=='x') { ncur.status[0]=0; ncur.log=0; } else { ncur.status[0]=ch-'0'; } for(int i=1;i<=8;i++) { cin>>ch; if(ch=='x') { ncur.status[i]=0; ncur.log=i; } else { ncur.status[i]=ch-'0'; } } ncur.hashd=cantor(ncur.status); if(BFS(ncur)) { stack<char> st; int id=aim; while(book[id].f!=0) { st.push(book[id].path); id=book[id].f; } while(!st.empty()) { cout<<st.top(); st.pop(); } } else cout<<"unsolvable"<<endl; } return 0; }

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