HDU 3567

Eight II

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 130000/65536K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 3

Problem Description

Eight-puzzle, which is also called "Nine grids", comes from an old game. 

In this game, you are given a 3 by 3 board and 8 tiles. The tiles are numbered from 1 to 8 and each covers a grid. As you see, there is a blank grid which can be represented as an 'X'. Tiles in grids having a common edge with the blank grid can be moved into that blank grid. This operation leads to an exchange of 'X' with one tile.

We use the symbol 'r' to represent exchanging 'X' with the tile on its right side, and 'l' for the left side, 'u' for the one above it, 'd' for the one below it.



A state of the board can be represented by a string S using the rule showed below.



The problem is to operate an operation list of 'r', 'u', 'l', 'd' to turn the state of the board from state A to state B. You are required to find the result which meets the following constrains:
1. It is of minimum length among all possible solutions.
2. It is the lexicographically smallest one of all solutions of minimum length.

 

Input

The first line is T (T <= 200), which means the number of test cases of this problem. The input of each test case consists of two lines with state A occupying the first line and state B on the second line. It is guaranteed that there is an available solution from state A to B.

 

Output

For each test case two lines are expected. The first line is in the format of "Case x: d", in which x is the case number counted from one, d is the minimum length of operation list you need to turn A to B. S is the operation list meeting the constraints and it should be showed on the second line.

 

Sample Input

   
   
   
   

    
    
    
    2
12X453786
12345678X
564178X23
7568X4123

   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Case 1: 2
dd
Case 2: 8
urrulldr

   
   
   
   

 

Author

zhymaoiing

 

Source

2010 ACM-ICPC Multi-University Training Contest（13）——Host by UESTC

 

和之前做的八数码差不多

因为X的位置不同  多出了9倍的状态  所以才去保存路径的方式预处理答案

因为题目保证都是合法情况  不需要判断逆序数

然后和八数码差不多思路  不需要A*  直接广搜预处理即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<string>
#define inf 1<<30
#define eps 1e-7
#define LD long double
#define LL long long
#define maxn 1000000005
using namespace std;
int ed;
int fac[3][3]= {{1,1,2},{6,24,120},{720,5040,40320}};//计算康拓值
struct node
{
    int Map[3][3];
    int x,y;
    int Hash;
    bool cheak()
    {
        if(x>=0&&x<3&&y>=0&&y<3)
            return true ;
        return false ;
    }
} u,uu,v;
int vis[9][400000];//标记
int pri[9][400000];//路径
int dir[4][2]= {{1,0},{0,-1},{0,1},{-1,0}};

int get_hash(int maze[3][3])  //计算康拓值
{
    int ret=0;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
        {
            int cnt=0;
            for(int a=0; a<i; a++)
                for(int b=0; b<3; b++)
                    if(maze[a][b]>maze[i][j])
                        cnt++;
            for(int b=0; b<j; b++)
                if(maze[i][b]>maze[i][j])
                    cnt++;
            ret+=cnt*fac[i][j];
        }
    return ret;
}

char ch[5]= {"dlru"};
//输出路径
void print(int i)
{
    int st=ed;

    int cnt=0,ans[1000];

    while(pri[i][st]!=-2)
    {
        ans[cnt++]=vis[i][st];
        st=pri[i][st];
    }
    printf("%d\n",cnt);
    for(int i=cnt-1; i>=0; i--)
        printf("%c",ch[ans[i]]);
    printf("\n");
    return ;
}

void bfs(int kind)
{
    queue<node>que;
    que.push(u);
    pri[kind][u.Hash]=-2;
    while(!que.empty())
    {
        uu=que.front();
        que.pop();
        for(int i=0; i<4; i++)
        {
            v=uu;
            v.x+=dir[i][0];
            v.y+=dir[i][1];
            if(v.cheak())
            {
                swap(v.Map[v.x][v.y],v.Map[uu.x][uu.y]);
                v.Hash=get_hash(v.Map);
                if(pri[kind][v.Hash]==-1)
                {
                    pri[kind][v.Hash]=uu.Hash;
                    vis[kind][v.Hash]=i;
                    que.push(v);
                }
            }
        }
    }
}

void Init(char *str,int kind)
{
    int k=0;
    for(int i=0; i<3; i++)
        for(int j=0; j<3; j++)
        {
            if(str[k]=='X')
                u.Map[i][j]=0;
            else
                u.Map[i][j]=str[k]-'0';
            k++;
        }
    u.x=kind/3;
    u.y=kind%3;
    u.Hash=get_hash(u.Map);
    bfs(kind);
}

int main()
{
    int t;
    int tt=1;
    //预处理
    memset(pri,-1,sizeof(pri));
    Init("X12345678",0);
    Init("1X2345678",1);
    Init("12X345678",2);
    Init("123X45678",3);
    Init("1234X5678",4);
    Init("12345X678",5);
    Init("123456X78",6);
    Init("1234567X8",7);
    Init("12345678X",8);
    scanf("%d",&t);
    while(t--)
    {
        char s1[10],s2[10];
        scanf("%s%s",s1,s2);
        int a[3][3],b[3][3],pos,c[9],k=1;
        for(int i=0; i<9; i++)
        {
            if(s1[i]=='X')
            {
                c[0]=0;
                pos=i;
            }
            else
                c[s1[i]-'0']=k++;
        }
        for(int i=0; i<9; i++)
            b[i/3][i%3]=(s2[i]=='X'?c[0]:c[s2[i]-'0']);
        ed=get_hash(b);
        printf("Case %d: ",tt++);
        print(pos);
    }
    return 0;
}