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HDU 3567

編輯：關於C++

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

Sample Output

Case 1: 2
dd
Case 2: 8
urrulldr

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

和之前做的八數碼差不多

因為X的位置不同多出了9倍的狀態所以才去保存路徑的方式預處理答案

因為題目保證都是合法情況不需要判斷逆序數

然後和八數碼差不多思路不需要A* 直接廣搜預處理即可

#include
#include
#include
#include
#include
#include
#include
#include
#include
#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}};//計算康拓值
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];//標記
int pri[9][400000];//路徑
int dir[4][2]= {{1,0},{0,-1},{0,1},{-1,0}};

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

char ch[5]= {dlru};
//輸出路徑
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
,cnt);
    for(int i=cnt-1; i>=0; i--)
        printf(%c,ch[ans[i]]);
    printf(
);
    return ;
}

void bfs(int kind)
{
    queueque;
    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;
    //預處理
    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;
}