Eight

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 a description of a configuration of the 8 puzzle. The 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.

Sample Input

 2  3  4  1  5  x  7  6  8 

Sample Output

ullddrurdllurdruldr
题意：给定一个二维图由其实状态得到目标状态输出最少步数的路径
题解：由起点开始搜索，单项BFS+hash。记录路径即可

#include<stdio.h>

#include<string.h>

#include<algorithm>

#include<iostream>

using namespace std;

int fan[9]={1,1,2,6,24,120,720,5040,40320};

int target[9]={1,2,3,4,5,6,7,8,0};

int sval,eval,po;

int visit[362881];

int c[9];

int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};

int mg[1000];

struct lmx{

    int a[9];

    int step;

    int pos;

    int pre;

    int dir;

};

void print(int i)

{

    if(i==0) printf("u");

    else if(i==1) printf("d");

    else if(i==2) printf("l");

    else if(i==3) printf("r");

}

int kgto(int c[],int n)

{

    int i,j,num,s=0;

    for(i=0;i<n-1;i++)

    {

        num=0;

        for(j=i+1;j<n;j++)

        {

            if(c[j]<c[i]) num++;

        }

        s+=num*fan[8-i];

    }

    return s;

}

lmx s,t,h,ss;

lmx lm[362881];

int bfs()

{

    visit[sval]=0;

    int head=-1,rear=0,i,xx,yy,val,x,y;

    memcpy(s.a,c,sizeof(c));

    s.step=0;

    s.pre=-1;

    s.dir=-1;

    s.pos=po;

    lm[head+1]=s;

    while(head<rear)

    {

        head++;

        h=lm[head];

        x=h.pos/3;

        y=h.pos%3;

        for(i=0;i<4;i++)

        {

            t=h;

            xx=x+dir[i][0];

            yy=y+dir[i][1];

            if(xx>=0&&xx<3&&yy>=0&&yy<3)

            {

                swap(t.a[h.pos],t.a[xx*3+yy]);

                t.pre=head;

                t.dir=i;

                val=kgto(t.a,9);

                t.pos=3*xx+yy;

                t.step=h.step+1;

                if(visit[val]==-1)

                {

                    if(val==eval) { ss=t;return t.step;}

                    lm[++rear]=t;

                    visit[val]=0;

                }

            }

        }

    }

    return -1;

}

int main()

{

    char mm[100],i,top;

    for(i=0;i<9;i++)

    {

        cin>>mm[i];

       if(mm[i]>='0'&&mm[i]<='8')  c[i]=mm[i]-'0';

       if(mm[i]=='x'){c[i]=0;po=i;}

    }

    eval=kgto(target,9);

    sval=kgto(c,9);

    memset(visit,-1,sizeof(visit));

    top=0;

    if(sval!=eval)

    {

        int temp=bfs();

        if(temp==-1) puts("unsolvable");

        else

        {

            while(ss.pre!=-1)

            {

                mg[top++]=ss.dir;

                ss=lm[ss.pre];

            }

            for(i=top-1;i>=0;i--)

            {

                print(mg[i]);

            }

            puts("");

        }

    }

    return 0;

}

解法2：

思路：由目标结点出发逆向打表记录路径

#include<stdio.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<queue>
#include<string.h>
using namespace std;
const int MAX=1000000;
int sr[9],po;
int fan[9]={1,1,2,6,24,120,720,5040,40320};
bool visit[MAX];
string ss[MAX];
struct lmx{
    int a[9];
    int pos;
    string str;
};
lmx s,t,h;
int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
char st[5]="durl";
int kgto(int m[],int n)
{
    int i,j,num,s=0;
    for(i=0;i<n-1;i++)
    {
        num=0;
        for(j=i+1;j<n;j++)
        {
            if(m[j]<m[i]) num++;
        }
        s+=num*fan[8-i];
    }
    return s;
}
int bfs()
{
    memset(visit,false,sizeof(visit));
    int i,val,x,y,xx,yy;
    queue<lmx>q;
    for(i=0;i<8;i++)
    {
        s.a[i]=i+1;
    }
    s.a[8]=0;
    s.pos=8;
    s.str="";
    q.push(s);
    val=kgto(s.a,9);
    visit[val]=true;
    ss[val]="";
    while(!q.empty())
    {
        h=q.front();
        q.pop();
        x=h.pos/3;
        y=h.pos%3;
        for(i=0;i<4;i++)
        {
            xx=x+dir[i][0];
            yy=y+dir[i][1];
            if(xx>=0&&xx<3&&yy>=0&&yy<3)
            {
                t=h;
                swap(t.a[h.pos],t.a[xx*3+yy]);
                t.pos=3*xx+yy;
                val=kgto(t.a,9);
                if(visit[val]==false)
                {
                    visit[val]=true;
                    t.str=st[i]+t.str;
                    q.push(t);
                    ss[val]=t.str;
                }
            }
        }
    }
}
int main()
{
    char ch;
    int i;
    bfs();
    while(cin>>ch)
    {
        if(ch>='0'&&ch<='9')  sr[0]=ch-'0';
        if(ch=='x') {sr[0]=0;po=0;}
        for(i=1;i<9;i++)
        {
            cin>>ch;
            if(ch>='0'&&ch<='9') sr[i]=ch-'0';
            else {sr[i]=0;po=i;}
        }
        int temp=kgto(sr,9);
        if(visit[temp]==false) puts("unsolvable");
        else cout<<ss[temp]<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/ffhuguang/p/3233033.html