HDU 1043 Eight (A* + HASH + 康托展开)

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13956    Accepted Submission(s): 3957 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

 

 

 

这题活生生搞了五天，为此学了A*和康托展开，虽然自己写出来了但还不是非常明白，有句话说“如果你真的懂了那么你就应该自己动手做一个”，看来我还没真的懂。

首先把每次矩阵的格局存结构里，然后对整个矩阵用康托展开来进行哈希，得到一个值，以此来判断此种情况是否出现过或是是否到达了终点，有一个很好的剪枝就是用奇偶逆序数来判断是否可解，因为最终情况的逆序数是0，而每次变换不会改变逆序数的奇偶，所以如果初始情况的逆序数是奇数的话就不可解。

不明白的地方是A*的估值函数为何不用f = g + h，而是要用h和g分别作为两个参数来比较，刚开始的时候我把两个参数的位置弄反了， T了一天。都是泪。

#include <iostream>
#include <cmath>
#include <string>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using    namespace    std;

const    int    SIZE = 5;
const    int    GOAL = 46233;
const    int    HASH[] = {40320,5040,720,120,24,6,2,1,1};
const    int    UP_DATE[][2] = {{0,-1},{0,1},{-1,0},{1,0}};
int    PATH[600000];
int    PRE[600000];
struct    Node
{
    int    map[SIZE][SIZE];
    int    x,y;
    int    h,g;
    int    hash;
    bool    operator <(const Node a) const
    {
        return    h != a.h ? h > a.h : g > a.g;
    }
};

bool    solve_able(const Node & r);
bool    check(const int,const int);
void    cal_hash(Node & r);
void    cal_h(Node & r);
void    search(const Node & r);
void    show(void);
int    main(void)
{
    Node    first;
    char    s[50];

    while(gets(s))
    {
        int    k = 0;
        memset(PRE,-1,sizeof(PRE));
        memset(PATH,-1,sizeof(PATH));
        for(int i = 1;i <= 3;i ++)
            for(int j = 1;j <= 3;j ++)
            {
                if(s[k] >= '1' && s[k] <= '9')
                    first.map[i][j] = s[k] - '0';
                else    if(s[k] == 'x')
                {
                    first.map[i][j] = 0;
                    first.x = i;
                    first.y = j;
                }
                else
                    j --;
                k ++;
            }
        if(!solve_able(first))
        {
            printf("unsolvable
");
            continue;
        }
        cal_hash(first);
        if(first.hash == GOAL)
        {
            puts("");
            continue;
        }
        PATH[first.hash] = -2;
        first.g = 0;
        cal_h(first);
        search(first);
    }

    return    0;
}

bool    solve_able(const Node & r)
{
    int    sum = 0,count = 0;
    int    temp[10];

    for(int i = 1;i <= 3;i ++)
        for(int j = 1;j <= 3;j ++)
        {
            temp[count] = r.map[i][j];
            count ++;
        }
    for(int i = 0;i < 9;i ++)
        for(int j = i + 1;j < 9;j ++)
            if(temp[j] < temp[i] && temp[j] && temp[i])
                sum ++;
    return    !(sum & 1);
}

bool    check(const int    x,const    int y)
{
    if(x >= 1 && x <= 3 && y >= 1 && y <= 3)
        return    true;
    return    false;
}

void    cal_hash(Node & r)
{
    int    sum = 0,count = 0,box;
    int    temp[10];

    for(int i = 1;i <= 3;i ++)
        for(int j = 1;j <= 3;j ++)
        {
            temp[count] = r.map[i][j];
            count ++;
        }
    for(int i = 0;i < 9;i ++)
    {
        box = 0;
        for(int j = i + 1;j < 9;j ++)
            if(temp[j] < temp[i])
                box ++;
        sum += (box * HASH[i]);
    }
    r.hash = sum;
}

void    search(Node const & r)
{
    Node    cur,next;

    priority_queue<Node>    que;
    que.push(r);
    while(!que.empty())
    {
        cur = que.top();
        que.pop();
        for(int i = 0;i < 4;i ++)
        {
            next = cur;
            next.x = cur.x + UP_DATE[i][0];
            next.y = cur.y + UP_DATE[i][1];
            if(!check(next.x,next.y))
                continue;
            swap(next.map[cur.x][cur.y],next.map[next.x][next.y]);
            cal_hash(next);

            if(PATH[next.hash] == -1)
            {
                PATH[next.hash] = i;
                PRE[next.hash] = cur.hash;
                next.g ++;
                cal_h(next);
                que.push(next);
            }
            if(next.hash == GOAL)
            {
                show();
                return    ;
            }
        }
    }

}

void    cal_h(Node & r)
{
    int    ans = 0;
    for(int i = 1;i <= 3;i ++)
        for(int j = 1;j <= 3;j ++)
            if(r.map[i][j])
                ans += abs(i - ((r.map[i][j] - 1) / 3 + 1)) + abs(j - ((r.map[i][j] - 1) % 3 + 1));
    r.h = ans;
}

void    show(void)
{
    string    ans;
    int    hash = GOAL;

    ans.clear();
    while(PRE[hash] != -1)
    {
        switch(PATH[hash])
        {
            case    0:ans += 'l';break;
            case    1:ans += 'r';break;
            case    2:ans += 'u';break;
            case    3:ans += 'd';break;
        }
        hash = PRE[hash];
    }
    for(int i = ans.size() - 1;i >= 0;i --)
        printf("%c",ans[i]);
    cout << endl;
}