HDU - 1043 Eight 打表+bfs+康托扩展

题目：

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：

因为最多只有9!种状态，一开始我想到的是直接暴力搜索起始状态到123456789x的状态，写完之后发现超时。

这里粘一下代码：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int maxn=362885;
char a[10];
int pos[4]={-3,3,-1,1};
char mea[6]="udlr";
char ans[maxn];
struct node
{
    int id;
    char a[10];
    int pre;
    int px;
    char op;
}b[maxn];
int cnt;
int cnt2;
setSet;
ll change (char s[])
{
    ll t=0;
    for (int i=0;i<9;i++)
    {
        t=t*10+(s[i]-'0');
    }
    return t;
}
int judge (ll t)
{
    if(Set.count(t)) return 0;
    Set.insert(t);
    return 1;
}
int bfs (int x)
{
    queueq;
    strcpy(b[0].a,a);
    b[0].pre=-1;
    b[0].px=x;
    b[0].op=-1;
    b[0].id=0;
    q.push(b[0]);
    cnt++;
    ll t=change(b[0].a);
    judge(t);
    node now,next;
    while(!q.empty())
    {
        now=q.front(); q.pop();
        ll t=change(now.a);
        if(t==123456780)
        {
            //printf("----\n");
            node x=now;
            while(x.pre!=-1)
            {
                ans[cnt2++]=x.op;
                x=b[x.pre];
            }
            return 1;
        }
        for (int i=0;i<4;i++)
        {
            next=now;
            int ch=now.px+pos[i];
            if((i==2&&now.px%3==0)||(i==3&&now.px%3==2)) continue;
            if(ch>=0&&ch<9)
            {
                swap(next.a[next.px],next.a[ch]);
                next.px=ch;
                ll t=change(next.a);
                if(judge(t))
                {
                    next.id=cnt;
                    next.pre=now.id;
                    next.op=i;
                    b[cnt++]=next;
                    q.push(next);
                }
            }
        }
    }
    return -1;
}
int main()
{
    while(scanf("%c ",&a[0])!=EOF)
    {
        int x;
        for (int i=1;i<8;i++) scanf("%c ",&a[i]);
        scanf("%c",&a[8]);
        for (int i=0;i<9;i++)
        {
            if(a[i]=='x')
            {
                a[i]='0';
                x=i;
            }
        }
        getchar();
        cnt=0; cnt2=0;
        Set.clear();
        if(bfs(x)!=-1)
        {
            for (int i=cnt2-1;i>=0;i--)
            {
                printf("%c",mea[ans[i]]);
            }
            printf("\n");
        }
        else
        {
            printf("unsolvable\n");
        }
    }
}

之后上网上搜了一下，发现应该从12345678x进行搜索，相当于进行打表，之后看看输入给出的状态是否可以从123456789x走到，这里需要储存走过的状态，网上普遍用了康托扩展，康托扩展确实使寻找状态快了很多，但是这个题目用set的话应该也能过，我感觉这个题目的侧重点应该是在打表上，康托扩展只是起到了锦上添花的效果。

这里贴一下AC代码，用了康托扩展。

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int maxn=362885;
int pos[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
//储存前驱节点
int can[maxn];
//储存方向
char op[maxn];
char a[50];
int f[10];
struct node
{
    int can; //现在节点的康托展开数
    int a[10];
    int x; //坐标
};
int cantor(int a[])
{
    int t=0;
    for (int i=0;i<9;i++)
    {
        int m=0;
        for (int j=i+1;j<9;j++)
        {
            if(a[i]>a[j]) m++;
        }
        t+=m*f[8-i];
    }
    return t;
}
void bfs ()
{
    queueq;
    node now,next;
    for (int i=0;i<9;i++) now.a[i]=i+1;
    now.x=8;
    now.can=0;
    can[0]=0;
    q.push(now);
    while(!q.empty())
    {
        now=q.front(); q.pop();
        for (int i=0;i<4;i++)
        {
            next=now;
            int tx=now.x/3+pos[i][0];
            int ty=now.x%3+pos[i][1];
            if(tx>=0&&tx<3&&ty>=0&&ty<3)
            {
                next.x=tx*3+ty;
                swap(next.a[tx*3+ty],next.a[now.x]);
                next.can=cantor(next.a);
                if(can[next.can]==-1)
                {
                    can[next.can]=now.can;
                    if(i==0) op[next.can]='l';
                    else if(i==1) op[next.can]='r';
                    else if(i==2) op[next.can]='u';
                    else if(i==3) op[next.can]='d';
                    q.push(next);
                }
            }
        }
    }
}
int main()
{
    f[0]=1;
    for (int i=1;i<=9;i++) f[i]=i*f[i-1];
    memset (can,-1,sizeof(can));
    bfs();
    while(gets(a)!=NULL)
    {
        int x,cnt=0;
        int len=strlen(a);
        int ca[10];
        for (int i=0;i'0'&&a[i]<'9') ca[cnt++]=a[i]-'0';
        }
        int t=cantor(ca);
        if(can[t]==-1) printf("unsolvable\n");
        else
        {
            while(t)
            {
                printf("%c",op[t]);
                t=can[t];
            }
            printf("\n");
        }
    }

}