POJ1077&&HDU1043(八数码，IDA*+曼哈顿距离)

Eight

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30127		Accepted: 13108		Special Judge

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 a description of a configuration of the 8 puzzle. The 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.

Sample Input

 2  3  4  1  5  x  7  6  8 

Sample Output

ullddrurdllurdruldr

//2016.8.25
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>

using namespace std;

int a[10][10], b[10], d[1000], sx, sy, deep;
bool ok;
map<int, bool> vis;
char dir[4] = {'u', 'd', 'l', 'r'};
int dx[4] = {-1, 1, 0, 0};//分别对应上下左右四个方向
int dy[4] = {0, 0, -1, 1};

bool solve()//求逆序对判断是否有解，偶数有解，奇数无解
{
    int cnt = 0;
    for(int i = 1; i <= 9; i++)
      for(int j = 0; j < i; j++)
        if(b[i] && b[j]>b[i])
          cnt++;
    return !(cnt%2);
}

int Astar()//启发函数，假设每个数字可以从别的数字头上跨过去，计算到达自己应该到的位置所需要的步数，即计算该数到达其正确位置的曼哈顿距离，h为各点曼哈顿距离之和
{
    int h = 0;
    for(int i = 1; i <= 3; i++)
        for(int j = 1; j <= 3; j++)
            if(a[i][j]!=0)
            {
                int nx = (a[i][j]-1)/3;
                int ny = (a[i][j]-1)%3;
                h += (abs(i-nx-1)+abs(j-ny-1));
            }
    return h;
}

int toInt()//把矩阵转换为int型数字
{
    int res = 0;
    for(int i = 1; i <= 3; i++)
      for(int j = 1; j <= 3; j++)
        res = res*10+a[i][j];
    return res;
}

void IDAstar(int x, int y, int step)
{
    if(ok)return ;
    int h = Astar();
    if(!h && toInt()==123456780)//找到答案
    {
        for(int i = 0; i < step; i++)
            cout<<dir[d[i]];
        cout<<endl;
        ok = 1;
        return ;
    }
    if(step+h>deep)return ;//现实＋理想<现状，则返回，IDA*最重要的剪枝
    int now = toInt();
    if(vis[now])return ;//如果状态已经搜过了，剪枝，避免重复搜索
    vis[now] = true;
    for(int i = 0; i < 4; i++)
    {
        int nx = x+dx[i];
        int ny = y+dy[i];
        if(nx>=1&&nx<=3&&ny>=1&&ny<=3)
        {
            d[step] = i;
            swap(a[x][y], a[nx][ny]);
            IDAstar(nx, ny, step+1);
            swap(a[x][y], a[nx][ny]);
            d[step] = 0;
        }
    }
    return;
}

int main()
{
    char ch;
    while(cin >> ch)
    {
        ok = false;
        deep = 0;
        int cnt = 0;
        for(int i = 1; i <= 3; i++)
        {
            for(int j = 1; j <= 3; j++)
            {
                if(i==1&&j==1);
                else cin >> ch;
                if(ch == 'x')
                {
                    a[i][j] = 0;
                    sx = i;
                    sy = j;
                }else
                  a[i][j] = ch - '0';
                b[cnt++] = a[i][j];
            }
        }
        if(!solve())
        {
            cout<<"unsolvable"<<endl;
            continue;
        }
        while(!ok)
        {
            vis.clear();
            IDAstar(sx, sy, 0);
            deep++;//一层一层增加搜索的深度
        }
    }

    return 0;
}