HDU3567

Eight II

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 130000/65536 K (Java/Others)
Total Submission(s): 3449    Accepted Submission(s): 753

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

 

//2017-09-23
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <map>

using namespace std;

const int INF = 0x3f3f3f3f;
int a[10][10], target, postion[10], b[10], d[10000], sx, sy, deep;
bool ok;
map<int, bool> vis;
char dir[4] = {'d', 'l', 'r', 'u'};
int dx[4] = {1, 0, 0, -1};//分别对应上下左右四个方向
int dy[4] = {0, -1, 1, 0};
int kase;

int Astar()
{
    int h = 0;
    for(int i = 1; i <= 3; i++)
      for(int j = 1; j <= 3; j++)
        if(a[i][j]!=0)
        {
            int num = a[i][j];
            int nx = (postion[num]-1)/3;
            int ny = (postion[num]-1)%3;
            h += (abs(i-nx-1)+abs(j-ny-1));
        }
    return h;
}

int toInt()//把矩阵转换为int型数字
{
    int res = 0;
    for(int i = 1; i <= 3; i++)
      for(int j = 1; j <= 3; j++)
        res = res*10+a[i][j];
    return res;
}

void IDAstar(int x, int y, int pre, int step)
{
    if(ok)return ;
    int h = Astar();
    if(!h && toInt()==target)//找到答案
    {
        printf("Case %d: %d\n", ++kase, step);
        for(int i = 0; i < step; i++)
          printf("%c", dir[d[i]]);
        printf("\n");
        ok = 1;
        return ;
    }
    if(step+h>deep)return ;//现实＋理想<现状，则返回，IDA*最重要的剪枝
    int now = toInt();
    //if(vis[now])return ;//如果状态已经搜过了，剪枝，避免重复搜索
    //vis[now] = true;
    for(int i = 0; i < 4; i++)
    {
        if(i+pre == 3)continue;
        int nx = x+dx[i];
        int ny = y+dy[i];
        if(nx>=1&&nx<=3&&ny>=1&&ny<=3)
        {
            d[step] = i;
            swap(a[x][y], a[nx][ny]);
            IDAstar(nx, ny, i, step+1);
            swap(a[x][y], a[nx][ny]);
            d[step] = 0;
        }
    }
    return;
}

char str1[20], str2[20];

int main()
{
    int T;
    scanf("%d", &T);
    char ch;
    kase = 0;
    while(T--)
    {
        ok = false;
        deep = 0;
        int cnt = 0;
        scanf("%s%s", str1, str2);
        for(int i = 1; i <= 3; i++)
        {
            for(int j = 1; j <= 3; j++)
            {
                ch = str1[(i-1)*3+j-1];
                if(ch == 'X')
                {
                    a[i][j] = 0;
                    sx = i;
                    sy = j;
                }else
                  a[i][j] = ch - '0';
                b[cnt++] = a[i][j];
            }
        }
        target = 0;
        getchar();
        for(int i = 1; i <= 9; i++){
            target *= 10;
            ch = str2[i-1];
            if(ch == 'X')
              target += 0;
            else{
                target += ch-'0';
                postion[ch-'0'] = i;
            }        
        }
        while(!ok)
        {
            vis.clear();
            IDAstar(sx, sy, INF, 0);
            deep++;//一层一层增加搜索的深度
        }
    }

    return 0;
}