[未完待续](c++实现)八数码Ⅱ

本题预计涉及:dfs搜索,康托展开(为了寻找映射来判重),曼哈顿距离

 

Eight II

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 130000/65536 K (Java/Others)
Total Submission(s): 8374    Accepted Submission(s): 1890

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

 

 

 

--------时间分界线----2021.4.19----

在初次应用了康托展开和dfs搜索后发现结果离答案还差一些(即输出的应是最短路径,而非寻找最短路径过程中走过的所有路径)

今天肝不动了,先把初次的代码安排上┭┮﹏┭┮呜呜

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define Judge(xx,yy) (xx>=0&&yy>=0&&xx<3&&yy<3)
#define Maxn 362880
using namespace std;

int start[9],goal[9];
int exist[Maxn];
char step[Maxn];
int factorial[]={1,1,2,6,24,120,720,5040,40320,362880}; 
struct Node{
    int num;//步数 
    int state[9];//状态 
};
int dir[4][2]={
    {1,0},//右 
    {0,1},//下 
    {-1,0},//左 
    {0,-1} //上 
}; 
bool Contor(int a[],int m)//康拓展开得到下标 
{
    long long site=0,cnt;//cnt记录比当前下标所在位更大的数的个数 
    for(int i=0; i<m; ++i)
    {
        cnt=0;
        for(int k=i+1; k<m; ++k)
        {
            if(a[i]<a[k]){
                 cnt++; 
            }
        }
        site+=cnt*factorial[m-i-1];//精髓 
    }
    if(!exist[site]){
        exist[site]=1;
        return true;
    }else{
        return false;
    }
}

int bfs()
{
    queue<Node> Q;
    Node head;
    head.num=0;
    memcpy(head.state,start,sizeof(head.state));
    Contor(head.state,9);
    
    Q.push(head);
    int foot=0;//步骤 step的下标 
    while(!Q.empty())
    {
        int zeroSite; 
        head=Q.front();
        if(!memcmp(goal,head.state,sizeof(goal)))
        {
            return head.num;
        }
        
        //调出X所在的一维坐标
        for(zeroSite=0;zeroSite<9;++zeroSite)
        {
            if(head.state[zeroSite]==0){
                break;
            }
        }
        Q.pop();
            
        int x=zeroSite%3;//第几列
        int y=zeroSite/3;//第几行
        int newZ;//移动后X所在的一维下标
        int newX,newY;

        for(int i=0; i<4; ++i)
        {
            newY=y+dir[i][1];
            newX=x+dir[i][0];
            newZ = 3*newY+newX;
            if(Judge(newX,newY)){
                Node next;
                memcpy(&next,&head,sizeof(struct Node));
                swap(next.state[zeroSite],next.state[newZ]);
                if(Contor(next.state,9)){
                    if(i==0){
                        step[foot]='r';
                    }else if(i==1){
                        step[foot]='d';
                    }else if(i==2){
                        step[foot]='l';
                    }else{
                        step[foot]='u';
                    }
                    foot++;
                    next.num++;
                    Q.push(next);
                }
            }
        }     
    }
    step[foot]='';
}
int main()
{
    char startCh[10],goalCh[10];
    int N;
    cin>>N;
    for(int i=1;i<=N;++i)
    {
        memset(exist,0,sizeof(int));
        memset(step,0,sizeof(char));
        scanf("%s",startCh);
        for(int j=0; j<9; ++j)
        {
            if(startCh[j]!='X'){
                start[j]=startCh[j]-'0';
            }else{
                start[j]=0;
            }
        }
        scanf("%s",goalCh);
        for(int j=0; j<9; ++j)
        {
            if(goalCh[j]!='X'){
                goal[j]=goalCh[j]-'0';
            }else{
                goal[j]=0;
            }
        }

        cout<<"Case"<<' '<<i<<':'<<' '<<bfs()<<endl;
        printf("%s",step);
    }
    return 0;
}

运行结果如下:

似乎还需要一些未掌握的方法.......(待续……)