Eight HDU-1043 （bfs）

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35625    Accepted Submission(s): 9219
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

 

思路：使用一维的string来表示当前局面，下标从0开始。

　　   但是代码超内存，可能是由于无解的情况导致的，待更新......

#include <iostream>
#include <queue>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <set>
#include <map>

using namespace std;

map<string, char> mp;    // 存储当前局面和方向
map<string, string>pre;    // 存储当前局面和上一局面

int flag = 0;

struct node
{
    int cur;       // x在string中的下标
    string s;    // 当前局面
}nod;

string Swap(string s, int x, int y)
{
    swap(s[x], s[y]);
    return s;
}

void Print(string str)  // 递归打印结果
{
    if(mp[str] == '#')
        return;
    Print(pre[str]);
    cout << mp[str];
}

void bfs()
{
    queue<node> Q;

    Q.push(nod);
    mp[nod.s] = '#';

    node p,t;
    while(!Q.empty())
    {
        p = Q.front();
        Q.pop();

        if(p.s == "12345678x")
        {
            flag = 1;
            Print("12345678x");
        }

        for(int i = 0; i < 4; ++i)
        {
            if(i == 3)    // 向左
            {
                if(p.cur % 3 != 0)  // 下标为0,3,6的不能向左移动
                {
                    t.s = Swap(p.s, p.cur, p.cur-1);
                    if(mp.count(t.s) == 0)
                    {
                        mp[t.s] = 'l';
                        pre[t.s] = p.s;
                        t.cur = p.cur - 1;
                        Q.push(t);
                    }

                }
            }
            else if(i == 2)    // 向右
            {
                if(p.cur % 3 != 2)  // 下标为2,5,8的不能向右移动
                {
                    t.s = Swap(p.s, p.cur, p.cur+1);
                    if(mp.count(t.s) == 0)
                    {
                        mp[t.s] = 'r';
                        pre[t.s] = p.s;
                        t.cur = p.cur + 1;
                        Q.push(t);
                    }

                }
            }
            else if(i == 1)    // 向上
            {
                if(p.cur > 2)   // 下标为0,1,2的不能向上移动
                {
                    t.s = Swap(p.s, p.cur, p.cur-3);
                    if(mp.count(t.s) == 0)
                    {
                        mp[t.s] = 'u';
                        pre[t.s] = p.s;
                        t.cur = p.cur - 3;
                        Q.push(t);
                    }

                }
            }
            else if(i == 0)    // 向下
            {
                if(p.cur < 6)   // 下标为6,7,8的不能向下移动
                {
                    t.s = Swap(p.s, p.cur, p.cur+3);
                    if(mp.count(t.s) == 0)
                    {
                        mp[t.s] = 'd';
                        pre[t.s] = p.s;
                        t.cur = p.cur + 3;
                        Q.push(t);
                    }
                }
            }

        }
    }
}


int main()
{

    char c;
    string str;
    int k;
    for(int i = 0; i < 9; ++i)
    {
        cin >> c;
        str += c;
        if(c == 'x')
            k = i;    // 记录x的初始下标
    }

    nod.s = str;
    nod.cur = k;

    bfs();
    if(flag == 0)
        cout << "unsolvable";
    cout << endl;

    return 0;
}