HDU 1043  和  POJ 1077   两题类似。。。但是输入不同。

HDU 上是同时多组输入，POJ是单组输入。

两个限时不同。

HDU 上反向搜索，把所有情况打表出来。

POJ上正向搜索。

这个题很经典，还需要继续做。先把第一次写的代码贴出来吧。

继续优化中

HDU 1043 

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7145    Accepted Submission(s): 1946
Special Judge

Problem 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, several descriptions of configuration of the 8 puzzle. One 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. Do not print a blank line between cases.  

Sample Input 2 3 4 1 5 x 7 6 8  

Sample Output ullddrurdllurdruldr  

Source
South Central USA 1998 (Sepcial Judge Module By JGShining)  

Recommend JGShining    

/*
HDU 1043 Eight
思路：反向搜索，从目标状态找回状态对应的路径
用康托展开判重


AC   G++  328ms  13924K

*/
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
#include<string>
using namespace std;
const int MAXN=1000000;//最多是9!/2
int fac[]={1,1,2,6,24,120,720,5040,40320,362880};//康拖展开判重
//         0!1!2!3! 4! 5!  6!  7!   8!    9!
bool vis[MAXN];//标记
string path[MAXN];//记录路径
int cantor(int s[])//康拖展开求该序列的hash值
{
    int sum=0;
    for(int i=0;i<9;i++)
    {
        int num=0;
        for(int j=i+1;j<9;j++)
          if(s[j]<s[i])num++;
        sum+=(num*fac[9-i-1]);
    }
    return sum+1;
}
struct Node
{
    int s[9];
    int loc;//“0”的位置
    int status;//康拖展开的hash值
    string path;//路径
};
int move[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//u,d,l,r
char indexs[5]="durl";//和上面的要相反，因为是反向搜索
int aim=46234;//123456780对应的康拖展开的hash值
void bfs()
{
    memset(vis,false,sizeof(vis));
    Node cur,next;
    for(int i=0;i<8;i++)cur.s[i]=i+1;
    cur.s[8]=0;
    cur.loc=8;
    cur.status=aim;
    cur.path="";
    queue<Node>q;
    q.push(cur);
    path[aim]="";
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
        int x=cur.loc/3;
        int y=cur.loc%3;
        for(int i=0;i<4;i++)
        {
            int tx=x+move[i][0];
            int ty=y+move[i][1];
            if(tx<0||tx>2||ty<0||ty>2)continue;
            next=cur;
            next.loc=tx*3+ty;
            next.s[cur.loc]=next.s[next.loc];
            next.s[next.loc]=0;
            next.status=cantor(next.s);
            if(!vis[next.status])
            {
                vis[next.status]=true;
                next.path=indexs[i]+next.path;
                q.push(next);
                path[next.status]=next.path;
            }
        }
    }

}
int main()
{
    char ch;
    Node cur;
    bfs();
    while(cin>>ch)
    {
        if(ch=='x') {cur.s[0]=0;cur.loc=0;}
        else cur.s[0]=ch-'0';
        for(int i=1;i<9;i++)
        {
            cin>>ch;
            if(ch=='x')
            {
                cur.s[i]=0;
                cur.loc=i;
            }
            else cur.s[i]=ch-'0';
        }
        cur.status=cantor(cur.s);
        if(vis[cur.status])
        {
            cout<<path[cur.status]<<endl;
        }
        else cout<<"unsolvable"<<endl;
    }
    return 0;
}

POJ  1077

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

Source

South Central USA 1998      

/*
POJ 1077 Eight
正向广度搜索
把“x"当初0

G++ AC 5200K 719ms


*/

#include<stdio.h>
#include<queue>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAXN=1000000;
int fac[]={1,1,2,6,24,120,720,5040,40320,362880};//康拖展开判重
//         0!1!2!3! 4! 5!  6!  7!   8!    9!
bool vis[MAXN];//标记

int cantor(int s[])//康拖展开求该序列的hash值
{
    int sum=0;
    for(int i=0;i<9;i++)
    {
        int num=0;
        for(int j=i+1;j<9;j++)
          if(s[j]<s[i])num++;
        sum+=(num*fac[9-i-1]);
    }
    return sum+1;
}
struct Node
{
    int s[9];
    int loc;//“0”的位置,把“x"当0
    int status;//康拖展开的hash值
    string path;//路径
};
string path;
int aim=46234;//123456780对应的康拖展开的hash值
int move[4][2]={{-1,0},{1,0},{0,-1},{0,1}};//u,d,l,r
char indexs[5]="udlr";//正向搜索
Node ncur;
bool bfs()
{
    memset(vis,false,sizeof(vis));
    Node cur,next;
    queue<Node>q;
    q.push(ncur);
    while(!q.empty())
    {
        cur=q.front();
        q.pop();
        if(cur.status==aim)
        {
            path=cur.path;
            return true;
        }
        int x=cur.loc/3;
        int y=cur.loc%3;
        for(int i=0;i<4;i++)
        {
            int tx=x+move[i][0];
            int ty=y+move[i][1];
            if(tx<0||tx>2||ty<0||ty>2)continue;
            next=cur;
            next.loc=tx*3+ty;
            next.s[cur.loc]=next.s[next.loc];
            next.s[next.loc]=0;
            next.status=cantor(next.s);
            if(!vis[next.status])
            {
                vis[next.status]=true;
                next.path=next.path+indexs[i];

                if(next.status==aim)
                {
                    path=next.path;
                    return true;
                }

                q.push(next);
            }
        }
    }
    return false;
}
int main()
{
    char ch;
    while(cin>>ch)
    {
        if(ch=='x') {ncur.s[0]=0;ncur.loc=0;}
        else ncur.s[0]=ch-'0';
        for(int i=1;i<9;i++)
        {
            cin>>ch;
            if(ch=='x')
            {
                ncur.s[i]=0;
                ncur.loc=i;
            }
            else ncur.s[i]=ch-'0';
        }
        ncur.status=cantor(ncur.s);
        if(bfs())
        {
            cout<<path<<endl;
        }
        else cout<<"unsolvable"<<endl;
    }
    return 0;
}

这个题目我的做法是把“x”当成0的。

网上很多当成9的话很多不一样了，特意说明下。

康托展开很简单，百度百科上的很容易理解。

谢谢

——————————kuangbin