HDU 1043 Eight

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16501    Accepted Submission(s): 4542
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

网上称这样的问题是八数码问题。对于八数码问题，有位大神表示八数码问题有八种境界，如果感兴趣可以点击这里。

在这里为了学习有关A*算法，所以使用的是A*算法+哈希+曼哈顿距离

哈希是根据康托展开变形作为建表基础的，表示的是当前位在八个数中的大小位置关系乘以所在位置的权值

简单介绍一下曼哈顿距离，即|x1 - x2| + |y1 - y2|，即在标准坐标系中是一个很好的启发式函数

对于A*算法其实就是一个介于BFS和dijkstra算法的启发式算法。有幸看到一份对A*算法介绍的文章，写的真的很棒，如果想深入理解的话点这里。

当然，这里我们根据A*算法思想，设置优先考虑启发式函数曼哈顿距离的参数，次优先考虑BFS的代价

另外值得一提的是，网上同样给出了，因为无论如何移动或者是交换位置，逆序的奇偶性就不会变，所以我们求出结果的逆序数，如果一开始或者在求解过程中出现逆序数的奇偶性的改变，我们就需要否定这个方案或者输出unsolvable。

代码如下：

/*************************************************************************
	> File Name: Eight_2.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Wed 07 Oct 2015 04:47:56 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <string>
#include <vector>
#include <cmath>

using namespace std;

struct Node{
    int Map[3][3];
    int h, g;
    int tx, ty;
    bool operator<(const Node x)const{
        return h != x.h ? h > x.h : g > x.g;
    }
    int Hash;
};

Node f;
bool check(int a, int b){
    if(a >= 0 && a < 3 && b >= 0 && b < 3)
        return true;
    else return false;
}

int Hash[9] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};

int Suc = 322560;

int vis[409115];
int Path[500000];

int di[4][2] = { {0, 1}, {0, -1}, {1, 0}, {-1, 0}};

int get_hash(Node x){
    int temp[9];
    int k = 0;

    for(int i = 0; i < 3; i ++){
        for(int j = 0; j < 3; j ++){
            temp[k ++] = x.Map[i][j];
        }
    }
    
    int res = 0;
    for(int i = 0; i < 9; i ++){
        k = 0;
        for(int j = 0; j < i; j ++){
            if(temp[j] > temp[i])
                k ++;
        }
        res += Hash[i] * k;
    }

    return res;
}

bool judge(Node x){
    int temp[9];
    int k = 0;
    for(int i = 0; i < 3; i ++){
        for(int j = 0; j < 3; j ++)
            temp[k ++] = x.Map[i][j];
    }
    int ans = 0;

    for(int i = 0; i < 9; i ++){
        for(int j = i + 1; j < 9; j ++){
            if(temp[j] && temp[i] && temp[i] > temp[j])
                ans ++;
        }
    }

    return !(ans & 1);
}

int get_h(Node x){
    int ans = 0;
    for(int i = 0; i < 3; i ++){
        for(int j = 0; j < 3; j ++){
            if(x.Map[i][j])
                ans += abs(i - (x.Map[i][j] - 1) / 3) + abs(j - (x.Map[i][j] - 1) % 3);       //曼哈顿距离
        }
    }
    return ans;
}

void bfs(){
    priority_queue<Node>q;
    Node temp, last;
    q.push(f);
    while(!q.empty()){
        last = q.top();
        q.pop();
        for(int i = 0; i < 4; i ++){
            temp = last;
            temp.tx += di[i][0];
            temp.ty += di[i][1];
            if(check(temp.tx, temp.ty)){
                swap(temp.Map[temp.tx][temp.ty], temp.Map[last.tx][last.ty]);
                temp.Hash = get_hash(temp);
                if(vis[temp.Hash] == -1 && judge(temp)){
                    vis[temp.Hash] = i;
                    temp.g ++;
                    Path[temp.Hash] = last.Hash;
                    temp.h = get_h(temp);
                    q.push(temp);
                }
                if(temp.Hash == Suc)
                    return ;
            }
        }
    }
}


void print(){
    string str;
    str.clear();
    int ok = Suc;
    while(Path[ok] != -1){
        if(vis[ok] == 0)
            str += 'r';
        else if(vis[ok] == 1)
            str += 'l';
        else if(vis[ok] == 2)
            str += 'd';
        else     
            str += 'u';
        ok = Path[ok];
    }
    for(int i = str.size() - 1; i >= 0; i --){
        putchar(str[i]);
    }
    puts("");
}


int main(void){
    char str[100];
    while(gets(str) != NULL){
        int k = 0;
        memset(vis, -1, sizeof(vis));
        memset(Path, -1, sizeof(Path));
        for(int i = 0; i < 3; i ++){
            for(int j = 0; j < 3; j ++){
                if((str[k] <= '9' && str[k] >= '0' || str[k] == 'x')){
                    if(str[k] == 'x'){
                        f.Map[i][j] = 0;
                        f.tx = i;
                        f.ty = j;
                    }
                    else
                        f.Map[i][j] = str[k] - '0';
                }
                else 
                    j --;
                k ++;
            }
        }
        if(!judge(f)){
            cout << "unsolvable"<< endl;
            continue;
        }
        f.Hash = get_hash(f);
        if(f.Hash == Suc){
            puts("");
            continue;
        }
        vis[f.Hash] = 1;
        f.g = 0;
        f.h = get_h(f);
        bfs();
        print();
    }
    return 0;
}