hdu 1043 Eight(八数码问题 高级搜索: A* 搜索)

Problem 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, 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
 
我是第一次做A*搜索 ,感觉好难啊,看了一天多了,表面的意思差不多明白了,但还没有深度的了解,以后慢慢再做这类题吧
是我的参考ACM之家上的题解,希望对你们有帮助

A*算法入门:http://www.cppblog.com/mythit/archive/2009/04/19/80492.aspx

本题详细解析请看:http://blog.csdn.net/acm_cxlove/article/details/7745323

第一个A*搜索,A*是一种启发式搜索,g为已花代价,h为估计的剩余代价,而A*是根据f=g+h作为估价函数进行排列,也就是优先选择可能最优的节点进行扩展。但是在搜索之前需要判断是否有解,而判断的方法就是根据逆序对数。
参考 代码
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include //常用数字操作 一般和algorithm搭配使用
#include //STL定义运算函数(代替运算符)
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF = 1000000000;
const int maxn =1000008;
struct Node
{
    int maze[3][3];
    int h,g;//估计函数
    int x,y;
    int Hash;
    bool operator< (const Node n1) const
    {
        return h!=n1.h? h>n1.h:g>n1.g;
    }
    bool check() //判断是否合法
    {
        if(x>=0&&x<3&&y>=0&&y<3)
            return true ;
        return false;
    }
};
Node s,u,v,tt;
int HASH[9]= {1,1,2,6,24,120,720,5040,40320};
int destination=322560; 
int vis[400000];
int pre[400000];
int way[4][2]= {{0,1},{0,-1},{1,0},{-1,0}};
//void debug(Node tmp)
//{
//    for(int i=0; i<3; i++)
//    {
//        for(int j=0; j<3; j++)
//            printf("%d ",tmp.maze[i][j]);
//        printf("\n");
//    }
//    printf("%d %d\n%d %d\n",tmp.x,tmp.y,tmp.g,tmp.h);
//    printf("hash=%d\n",tmp.Hash);
//}
int get_hash(Node tmp)
{
    int a[9],k=0;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
            a[k++]=tmp.maze[i][j];
    int res=0;
    for(int i=0; i<9; i++)
    {
        int k=0;
        for(int j=0; ja[i]) k++;
        res+=HASH[i]*k;
    }
    return res;
}
bool isok(Node tmp)
{
    int a[9],k=0;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
            a[k++]=tmp.maze[i][j];
    int sum=0;
    for(int i=0; i<9; i++)
        for(int j=i+1; j<9; j++)
            if(a[j]&&a[i]&&a[i]>a[j]) sum++;
    return !(sum&1);
}
int get_h(Node tmp)
{
    int ans=0;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
            if(tmp.maze[i][j])
                ans+=abs(i-(tmp.maze[i][j]-1)/3)+abs(j-(tmp.maze[i][j]-1)%3);
    return ans;
}
void astar()
{
    priority_queueque;
    que.push(s);
    while(!que.empty())
    {
        u=que.top();
        que.pop();
        for(int i=0; i<4; i++)
        {
            v=u;
            v.x+=way[i][0];
            v.y+=way[i][1];
            if(v.check())
            {
                swap(v.maze[v.x][v.y],v.maze[u.x][u.y]);
                v.Hash=get_hash(v);
                if(vis[v.Hash]==-1&&isok(v))
                {
                    vis[v.Hash]=i;
                    v.g++;
                    pre[v.Hash]=u.Hash;
                    v.h=get_h(v);
                    que.push(v);
                }
                if(v.Hash==destination) return;
            }
        }
    }
}
void print()
{
    string ans;
    ans.clear();
    int nxt=destination;
    while(pre[nxt]!=-1)
    {
        switch(vis[nxt])
        {
        case 0:
            ans+='r';
            break;
        case 1:
            ans+='l';
            break;
        case 2:
            ans+='d';
            break;
        case 3:
            ans+='u';
            break;
        }
        nxt=pre[nxt];
    }
    for(int i=ans.size()-1; i>=0; i--)
        putchar(ans[i]);
    puts("");
}
int main()
{
    char str[100];
    while(gets(str)!=NULL)
    {
        int k=0,i,j;
        memset(vis,-1,sizeof(vis));
        memset(pre,-1,sizeof(pre));
        for(i=0; i<3; i++)
        {
            for(j=0; j<3; j++)
            {
                if((str[k]<='9'&&str[k]>='0')||str[k]=='x')
                {
                    if(str[k]=='x')
                    {
                        s.maze[i][j]=0;
                        s.x=i;
                        s.y=j;
                    }
                    else
                        s.maze[i][j]=str[k]-'0';
                }
                else
                j--;
                k++;
            }
        }
        if(!isok(s))//起始状态不可行
        {
            printf("unsolvable\n");
            continue;
        }
        s.Hash=get_hash(s);
        if(s.Hash==destination)//起始状态为目标状态
        {
            puts("");
            continue;
        }
        vis[s.Hash]=-2;
        s.g=0;
        s.h=get_h(s);
        astar();
        print();
    }
    return 0;
}







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