hdu1043 bfs 康拓展开

http://acm.hdu.edu.cn/showproblem.php?pid=1043

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
   
   
   
   

/**
hdu1043 bfs
题目大意：类似一个九宫格问题，给定一个3*3的棋盘，有1~8个数字，一个x。给定初始位置状态，最终到达同一指定状态，x棋子可以和其上下左右的棋子进行交换顺序，求出一种给定的合法序列
          从初始装太可以到达给定状态
解题思路：基本的方法是利用康拓展开记录状态，从初始状态到给定状态进行bfs搜索。但是这样写有多少组输入就要进行多少次bfs比较费时。因为题目给定的是同一个目标状态，所以我们进行倒着搜
          搜索出所有的状态，这样只进行一次bfs就可以了
*/
#include <string.h>
#include <algorithm>
#include <string>
#include <iostream>
#include <stdio.h>
#include <queue>
using namespace std;
const int maxn=1e6+5;
const int INF=0x3f3f3f3f;

struct note
{
    char s[3][3];
    int x,y;
    int can;
};

int father[maxn],path[maxn],vis[maxn];
int dx[]= {-1,1,0,0};
int dy[]= {0,0,-1,1};
int fac[10];

void init()
{
    fac[0]=1;
    fac[1]=1;
    for(int i=2; i<10; i++)
    {
        fac[i]=fac[i-1]*i;
    }
    memset(vis,0,sizeof(vis));
    memset(father,-1,sizeof(father));
}

int canton(char s[3][3])
{
    int a[9];
    int cur=0;
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
        {
            a[cur++]=s[i][j]-'0';
        }
    }
    int ans=0;
    for(int i=0; i<9; i++)
    {
        int cnt=0;
        for(int j=i+1; j<9; j++)
        {
            if(a[j]<a[i])cnt++;
        }
        ans+=cnt*fac[8-i];
    }
    return ans;
}

void bfs(note ss)
{
    queue<note>q;
    int ca=canton(ss.s);
    vis[ca]=1;
    father[ca]=-2;
    ss.can=ca;
    q.push(ss);
    while(!q.empty())
    {
        note st=q.front(),en;
        q.pop();
        for(int i=0; i<4; i++)
        {
            int x=st.x+dx[i];
            int y=st.y+dy[i];
            if(x>=0&&x<3&&y>=0&&y<3)
            {
                en=st;
                swap(en.s[x][y],en.s[st.x][st.y]);
                int t=canton(en.s);
                if(!vis[t])
                {
                    vis[t]=1;
                    en.x=x;
                    en.y=y;
                    en.can=t;
                    q.push(en);
                    father[t]=st.can;
                    path[t]=i;
                }
            }
        }
    }
}

void print(int x)
{
    if(father[x]==-2)return;
    int t=path[x];
    if(t==0)putchar('d');
    else if(t==1)putchar('u');
    else if(t==2)putchar('r');
    else putchar('l');
    print(father[x]);
}

int main()
{
    init();
    note s;
    int cnt=1;
    for(int i=0; i<3; i++)
    {
        for(int j=0; j<3; j++)
        {
            s.s[i][j]=cnt+'0',cnt++;
        }
    }
    s.s[2][2]='0';
    s.x=s.y=2;
    bfs(s);
    char ss[3][3];
    char sss[100];
    while(gets(sss)!=NULL)
    {
        int cur=0;
        for(int i=0; i<3; i++)
        {
            for(int j=0; j<3; j++)
            {
                while(sss[cur]==' '&&sss[cur]!='\0')cur++;
                if(sss[cur]<='9'&&sss[cur]>='0')
                    ss[i][j]=sss[cur++];
                else if(sss[cur]=='x')
                {
                    ss[i][j]='0';
                    cur++;
                }
            }
        }
        int sta=canton(ss);
        if(father[sta]==-1)
            printf("unsolvable");
        else
            print(sta);
        cout<<endl;
    }
    return 0;
}