搜索专题训练hdu1043Eight

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

经典的八数码问题，类似于我们平时玩的拼图。在解题过程中遇到了许多问题，参考了别人的做法，总结整理一下。

首先对于一个图的状态，我们要将其对应为一个status，通过康托展开，将其映射到hash空间。 康托展开利用了从高位到低位，每一位中比其小并且在高位中没有出现过的数字的个数，乘以该数位的阶乘。（我们可以通过这种方式得到比他小的数有多少个）。

把每一种情况hash映射以后，我们就要对当前状态到目标状态进行搜索，为了使路径尽可能短并且搜索更高效，首先想到了bfs，即从给定的状态，每次将x移动一步进行广度搜索，结果一定是TLE

由于测试是special judge而且是多组数据，想到离线bfs，即从终点开始，反向搜索，找出所有的可能解，然后根据给定状态是否被访问，直接输出答案，结果内存超限，bfs的层数太多了。

同时通过参考得知，无解的情况是可以直接判断来剪枝的，由于目标街12345678x的逆序数是0，其中x看做空位，每次变换就是x和某一位数交换，由于是空位交换，可以看做两次普通的数字交换，我们知道一次任意数字变换必定改变序列的逆序奇偶性，所以x和数字交换一定不改变奇偶性，所以可以直接判断逆序奇偶性来判断是否有解。

最后采用A*的方法，其中f = g + h  g用起始到当前的步数决定，h由当前状态到目标状态的曼哈顿距离表示。在搜索过程中，路径以记录上一节点位置和移动方向来记录，同时得到的结果是反向的，反向输出结果即可。

#include<bits/stdc++.h>using namespace std;const int MAXN=1000000;//最多是9!int fac[]={1,1,2,6,24,120,720,5040,40320,362880};//康拖展开判重bool vis[MAXN];//标记char path[MAXN];int pre[MAXN];int cantor(int s[])//康拖展开求该序列的hash值{    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+1;}struct Node{    int s[9];    int f, g, h; //存储函数    int loc;//“0”的位置    int status;//康拖展开的hash值    friend bool operator< (const Node &a,const Node &b)    {        if(a.f==b.f)            return a.g<b.g;        return a.f>b.f;    }};int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//u,d,l,rchar indexs[5]="udlr";int aim=46234;//123456780对应的康拖展开的hash值int cheak(int s[]){    int cnt = 0;    for(int i = 0; i < 9; ++i)    {        for(int j = i+1; j < 9; ++j)            if(s[j]<s[i] && s[j] && s[i]) cnt++;    }    if(cnt%2)        return 0;    return 1;}int get_h(const Node &x){    int ans=0;    for(int i=0;i<9;i++)    {        if(x.s[i] == 0) continue;        ans += abs(i/3-(x.s[i]-1)/3)+abs(i%3-(x.s[i]-1)%3);    }    //cout << ans << endl;    return ans;}void bfs(Node &cur){    //cout << "ok" << endl;    memset(vis,false,sizeof(vis));    Node next;    cur.g = 0;    cur.h = get_h(cur);    cur.f = cur.g + cur.h;    priority_queue<Node> q;    q.push(cur);    while(!q.empty())    {        cur=q.top();        q.pop();        int x=cur.loc/3;        int y=cur.loc%3;        for(int i=0;i<4;i++)        {            int tx=x+dir[i][0];            int ty=y+dir[i][1];            if(tx<0||tx>2||ty<0||ty>2)continue;            next=cur;            next.loc=tx*3+ty;            next.s[cur.loc]=next.s[next.loc];            next.s[next.loc]=0;            next.status=cantor(next.s);            next.g++;            next.h = get_h(next);            next.f = next.g + next.h;            if(!vis[next.status])            {                vis[next.status]=true;                pre[next.status] = cur.status;                path[next.status]= indexs[i];                q.push(next);                if(next.status == aim)                    return;            }        }    }}int main(){    char ch;    Node cur;    while(cin>>ch)    {        if(ch=='x') {cur.s[0]=0;cur.loc=0;}        else cur.s[0]=ch-'0';        for(int i=1;i<9;i++)        {            cin>>ch;            if(ch=='x')            {                cur.s[i]=0;                cur.loc=i;            }            else cur.s[i]=ch-'0';        }        if(!cheak(cur.s))        {            cout << "unsolvable" << endl;            continue;        }        cur.status=cantor(cur.s);        int start = cur.status;        bfs(cur);        int now = aim;        stack<char> st;        while(start != now)        {            st.push(path[now]);            now = pre[now];        }        while(!st.empty())        {            cout << st.top();            st.pop();        }        cout << endl;    }    return 0;}