八数码

时间：2015-07-23 15:25:26 阅读：141 评论：0 收藏：0 [点我收藏+]

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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set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;

}

八数码

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原文地址：http://www.cnblogs.com/zsyacm666666/p/4670491.html

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