HDU 1043 八数码 Eight A*算法

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10101    Accepted Submission(s): 2684
Special Judge

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

 

#include <iostream>
#include <string>
#include <string.h>
#include <map>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <vector>
#include <math.h>
#include <set>
#define Max(a,b) ((a)>(b)?(a):(b))
#pragma comment(linker, "/STACK:16777216")
using namespace std ;
typedef long long LL ;
const int fac[] = {,,,,,,,,} ;
const int d[][] = {{,},{-,},{,},{,-}} ; //四个方向
const char direction[] = {'d','u','r','l'} ;     //输出记录
const int end_pos[][] = {{,},{,},{,},{,},
                                 {,},{,},{,},
                                 {,},{,},{,}}  ;  //最后状态  123
                                                        //          456
                                                        //          78X

class Node{
    public :
        char mat[][] ;
        int  x ;
        int  y ;
        int  h ;  //h(x)=递归层数
        int  g ;  //g(x)=曼哈顿距离和
        int  f ;  //估价函数 f(x) = g(x)+ h(x)
        int  id ; //hash值,用于判重,利用排列序列的康拓展开
        int  G() ;//求g（x）
        int  cango() ; //剪枝，为什么逆序数必须是偶数，网上很多人问这个问题。
                       //我的解答： 与X（空白）交换的Y ，编号（index）相差为偶数，即在[ index[X] + 1 .index[Y]-1 ] 区间的数为偶数 ；
                       // 交换前  ，设  [ index[X] + 1 .index[Y]-1 ](zh) ，〉Y有a个，<Y 有b个 ，则a+b为偶数 ；
                       // 交换后  ，逆序数相较之前 +b -a  ,即| 逆序数交换前 -逆序数交换后 | 为偶数 。
                       // 也就是说无论何种交换，逆序数的变化差值为偶数 ，而最终的状态12345678X   。 (X可看作为9) ，逆序数为0。
                       // 即只能从逆序数为偶数的状态转移  。
        int  Hash() ;  // 求id
        void out() ;
        Node(){};
        Node(vector<char>) ;
        friend bool operator < (const Node &A ,const Node &B){
             if(A.f != B.f)
                return A.f > B.f ;
             else
                return A.g > B.g ;
        }             //建堆，使用C++ STL 优先队列 ，f(x)=g(x)+h（x） , f(x)小的优先级别大，也就是说与最后的状态差异小的优先考虑
};

Node::Node(vector<char>s){
    int i , j ;
    for(int k =  ;k < s.size() ;k++){
       i = k/ ;
       j = k% ;
       if(s[k] == 'x'){
           this->x = i ;
           this->y = j ;
       }
       mat[i][j] = s[k] ;
    }
}

int Node::G(){
    int sum =  ;
    for(int i =  ;i <=  ;i++){
      for(int j =  ;j <=  ;j++){
         if(mat[i][j]=='x')
           continue ;
         int n = mat[i][j] - '' ;
         sum = sum + abs(i-end_pos[n][]) + abs(j - end_pos[n][]) ;
      }
    }
    return  sum ;
}

int Node::cango(){
    char num[] ;
    int n =  ,sum = ;
    for(int i =  ;i <= ;i++){
        for(int j =  ;j <=  ;j++){
            if(mat[i][j] != 'x')
               num[++n] = mat[i][j] ;
        }
    }
    for(int i =  ;i <= n ;i++){
        for(int j = ;j < i ;j++){
            if(num[j] > num[i])
               sum++ ;
        }
    }
    return (sum&) ==  ;
}

int Node::Hash(){
    char num[] ;
    int n =  ,sum ,index =  ;
    for(int i =  ;i <= ;i++){
        for(int j =  ;j <=  ;j++)
               num[n++] = mat[i][j] ;
    }
    for(int i =  ;i < n ;i++){
        sum =  ;
        for(int j = ;j < i ;j++){
            if(num[j] > num[i])
               sum++ ;
        }
        index += fac[i]*sum ;
    }
    return  index ;
}

void Node::out(){
    for(int i =  ;i <=  ;i++){
        for(int j =  ;j <=  ;j++)
            putchar(mat[i][j]) ;
        puts("") ;
    }
}

int father[]  ;
class App{
  private :
    bool visited[] ;
    Node start ;
    char opet[] ;
  public :
    App(){}
    App(class Node) ;
    int cango(int ,int) ;
    int A_star() ;
    void out() ;
};

App::App(class Node s){
    start = s ;
    start.h =  ;
    start.g = start.G() ;
    start.f = start.g + start.h ;
    start.id = start.Hash() ;
    memset(visited,,sizeof(visited)) ;
    visited[start.id] =  ;
}

int App::cango(int x ,int y){
    return <=x&&x<=&&<=y&&y<= ;
}

int App::A_star(){
    if(start.id == )
        return  ;
    if(!start.cango())
        return  ;
    priority_queue<Node>que ;
    que.push(start) ;
    while(!que.empty()){
        Node now =que.top() ;
        que.pop() ;
        if(now.id == )
           return  ;
        for(int i =  ;i <  ;i++){
           int x = now.x + d[i][] ;
           int y = now.y + d[i][] ;
           if(!cango(x,y))
             continue ;
           Node next = now ;
           next.x = x ;
           next.y = y ;
           swap(next.mat[now.x][now.y],next.mat[next.x][next.y]) ;
           next.id = next.Hash() ;
           if(visited[next.id])
              continue ;
           visited[next.id] =  ;
           if(!next.cango())
              continue ;
           next.g = next.G() ;
           next.h = now.h +  ;
           next.f = next.g + next.h ;
           father[next.id] = now.id ;
           opet[next.id] = direction[i] ;
           que.push(next) ;
        }
    }
    return  ;
}

void App::out(){
    if(A_star()==)
       puts("unsolvable") ;
    else{
        vector<char>ans ;
        ans.clear() ;
        int i =  ;
        while(start.id != i){
            ans.push_back(opet[i]) ;
            i = father[i] ;
        }
        for(int i = ans.size()- ;i >=  ;i--)
           putchar(ans[i]) ;
        puts("") ;
    }
}

int main(){
   char str[] ;
   vector<char> s ;
   while(gets(str)){
       s.clear() ;
       for(int i =  ;i < strlen(str) ;i++){
           if(str[i] != ' ')
               s.push_back(str[i]) ;
       }
       Node start ;
       start = Node(s) ;
       App app = App(start)  ;
       app.out() ;
   }
   return  ;
}

相关知识点：

A*算法

公式表示为： f(n)=g(n)+h(n),

其中 f(n) 是从初始点经由节点n到目标点的估价函数，

g(n) 是在状态空间中从初始节点到n节点的实际代价，

h(n) 是从n到目标节点最佳路径的估计代价。

保证找到最短路径（最优解的）条件，关键在于估价函数h(n)的选取：

估价值h(n)<= n到目标节点的距离实际值，这种情况下，搜索的点数多，搜索范围大，效率低。但能得到最优解。

如果 估价值>实际值,搜索的点数少，搜索范围小，效率高，但不能保证得到最优解。

康托展开

 

{1,2,3,4,...,n}表示1,2,3,...,n的排列如 {1,2,3} 按从小到大排列一共6个 123 132 213 231 312 321

代表的数字 1 2 3 4 5 6 也就是把10进制数与一个排列对应起来。他们间的对应关系可由康托展开来找到。 如我想知道321是{1,2,3}中第几个大的数可以这样考虑 第一位是3，当第一位的数小于3时，那排列数小于321 如 123 213 小于3的数有1，2 所以有2*2!个 再看小于第二位2的 小于2的数只有一个就是1 所以有1*1!=1 所以小于321的{1,2,3}排列数有2*2!+1*1!=5个所以321是第6个大的数。 2*2!+1*1!是康托展开 再举个例子 1324是{1,2,3,4}排列数中第几个大的数 第一位是1小于1的数没有，是0个 0*3! 第二位是3小于3的数有1，2但1已经在第一位了所以只有一个数2 1*2! 第三位是2小于2的数是1，但1在第一位所以有0个数 0*1! 所以比1324小的排列有0*3!+1*2!+0*1!=2个 1324是第三个大数。