hdu 1043 /poj 1077 Eight（经典八数码问题，BFS+康托展开）

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20170    Accepted Submission(s): 5371
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

 

Source

South Central USA 1998 (Sepcial Judge Module By JGShining)

 

题意：经典八数码问题，移动x使得序列变成12345678_（_是空格），求最少移动次数的一个序列

思路：这个题hdu比poj的要紧一点，所以hdu能过的话poj一般就没问题。

用bfs去打表，一共只有9！（大概37w）个状态，不算多，记录从12345678_这个状态能够走到的所有的状态，并记录其父亲状态以及走的方向

然后我们每次输入序列只要一直找父亲，边输出路径，直到找到根结束即可。

方向就记反向的，方便我们找的时候直接输出。

记录一个序列，我们可以用hash，当然用普通hash是不行的，要用到康托展开hash，可以去看下我的前一篇博客或者百度百科，很好理解的东西。

不太会用A*...所以还没看懂A*的解法，真是菜啊=-=

代码：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
struct Node
{
    char step;
    int father;
} node[380005];
struct Node1
{
    int a[10];
    int n,id;
};
int f[10],num[10];
int dir[4][2]= {-1,0,1,0,0,1,0,-1};
void init()
{
    f[0]=1;
    for(int i=1; i<=8; i++)
        f[i]=f[i-1]*i;
    for(int i=1; i<380000; i++)
        node[i].father=-1;
    node[0].father=0;
}
int cantor(int *a)
{
    int ans=0;
    for(int i=1; i<=9; i++)
    {
        int k=0;
        for(int j=i+1; j<=9; j++)
            if(a[i]>a[j])
                k++;
        ans+=k*f[9-i];
    }
    return ans;
}
void bfs()
{
    queueque;
    Node1 q,next;
    memcpy(q.a,num,sizeof(num));
    q.n=9;
    q.id=0;
    que.push(q);
    while(!que.empty())
    {
        Node1 now=que.front();
        que.pop();
        for(int i=0; i<4; i++)
        {
            int x= now.n%3==0?now.n/3+dir[i][0]:now.n/3+1+dir[i][0];
            int y= now.n%3==0?dir[i][1]+3:now.n%3+dir[i][1];
            if(x<1||x>3||y<1||y>3) continue;
            next.n=(x-1)*3+y;
            memcpy(next.a,now.a,sizeof(now.a));
            swap(next.a[now.n],next.a[next.n]);
            next.id=cantor(next.a);
            if(node[next.id].father==-1)
            {
                node[next.id].father=now.id;
                if(i==0) node[next.id].step='d';
                else if(i==1) node[next.id].step='u';
                else if(i==2) node[next.id].step='l';
                else node[next.id].step='r';
                que.push(next);
            }
        }
    }
}
int main()
{
    char op[100];
    int ans[10];
    init();
    for(int i=1; i<=9; i++)
        num[i]=i;
    bfs();
    while(gets(op)>0)
    {
        int t=0;
        for(int i=0; i='0'&&op[i]<='9') ans[++t]=op[i]-'0';
        }
        int id=cantor(ans);
        if(node[id].father==-1)
        {
            printf("unsolvable\n");
            continue;
        }
        while(id)
        {
            printf("%c",node[id].step);
            id=node[id].father;
        }
        printf("\n");
    }
    return 0;
}