• hdu 1043 八数码问题


    Eight

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 10778    Accepted Submission(s): 2873
    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
     
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     康托展开优化,代码自己写的过了,做EIGHTII的时候,发现别人bfs()很简单,而且map[4][2]写的看不懂。
     
      1 #include<iostream>
      2 #include<stdio.h>
      3 #include<cstring>
      4 #include<cstdlib>
      5 #include<string>
      6 #include<queue>
      7 using namespace std;
      8 
      9 struct node
     10 {
     11     bool flag;
     12     char str;
     13     int father;
     14 }hash[363001];
     15 struct st
     16 {
     17     char t[10];
     18 };
     19 queue<st>Q;
     20 int ans[10]={1};
     21 
     22 int ktzk(char *c)
     23 {
     24     int i,j,k,sum=0;
     25     for(i=0;i<=8;i++)
     26     {
     27         k=0;
     28         for(j=i+1;j<=8;j++)
     29             if(c[i]>c[j])
     30                 k++;
     31         sum=sum+k*ans[8-i];
     32     }
     33     return sum;
     34 }
     35 void bfs()
     36 {
     37     int i,k,num,val;
     38     struct st cur,t;
     39     k=ktzk("123456780");
     40     hash[k].flag=true;
     41     hash[k].str='';
     42     hash[k].father=k;
     43 
     44     strcpy(t.t,"123456780");
     45     Q.push(t);
     46 
     47     while(!Q.empty())
     48     {
     49         cur=Q.front();
     50         Q.pop();
     51         val=ktzk(cur.t);
     52         for(i=0;i<=8;i++)
     53         {
     54             if(cur.t[i]=='0')
     55             {
     56                 k=i;
     57                 break;
     58             }
     59         }
     60         if(k!=2 && k!=5 && k!=8)//rigth
     61         {
     62             t=cur;
     63             swap(t.t[k],t.t[k+1]);
     64             num=ktzk(t.t);
     65             if(hash[num].flag==false)
     66             {
     67                 hash[num].flag=true;
     68                 hash[num].str='r';
     69                 hash[num].father=val;
     70                 Q.push(t);
     71             }
     72         }
     73         if(k!=0 && k!=3 && k!=6)//left
     74         {
     75             t=cur;
     76             swap(t.t[k],t.t[k-1]);
     77             num=ktzk(t.t);
     78             if(hash[num].flag==false)
     79             {
     80                 hash[num].flag=true;
     81                 hash[num].str='l';
     82                 hash[num].father=val;
     83                 Q.push(t);
     84             }
     85         }
     86         if(k>=3)//u
     87         {
     88             t=cur;
     89             swap(t.t[k],t.t[k-3]);
     90             num=ktzk(t.t);
     91             if(hash[num].flag==false)
     92             {
     93                 hash[num].flag=true;
     94                 hash[num].str='u';
     95                 hash[num].father=val;
     96                 Q.push(t);
     97             }
     98         }
     99         if(k<=5)//D
    100         {
    101             t=cur;
    102             swap(t.t[k],t.t[k+3]);
    103             num=ktzk(t.t);
    104             if(hash[num].flag==false)
    105             {
    106                 hash[num].flag=true;
    107                 hash[num].str='d';
    108                 hash[num].father=val;
    109                 Q.push(t);
    110             }
    111         }
    112     }
    113 }
    114 void prepare()
    115 {
    116     int i;
    117     for(i=0;i<=363000;i++)    
    118     {
    119         hash[i].flag=false;
    120     }
    121     for(i=1;i<=9;i++) ans[i]=ans[i-1]*i;
    122     bfs();
    123 }
    124 int main()
    125 {
    126     prepare();
    127     char a[50],b[10],c[100];
    128     bool tom[10];
    129     int i,j,k,cur;
    130     while(gets(a))
    131     {
    132         k=strlen(a);
    133         b[9]='';
    134         for(i=0,j=0;i<k;i++)
    135         {
    136             if(a[i]=='x' || (a[i]>='1'&&a[i]<='8'))
    137             {
    138                 if(a[i]=='x')
    139                     b[j]='0';
    140                 else b[j]=a[i];
    141                 j++;
    142             }
    143         }
    144         memset(tom,false,sizeof(tom));
    145         for(i=0;i<=8;i++)
    146         {
    147             tom[b[i]-'0']=true;
    148         }
    149         for(i=0;i<=8;i++)
    150         {
    151             if(tom[i]==false)
    152                 break;
    153         }
    154         if(i<=8){printf("unsolvable
    ");continue;}
    155         cur=ktzk(b);
    156         if(hash[cur].flag==false)
    157         {
    158             printf("unsolvable
    ");
    159             continue;
    160         }
    161         k=0;
    162         while(hash[cur].father!=cur)
    163         {
    164             c[k]=hash[cur].str;
    165             k++;
    166             cur=hash[cur].father;
    167         }
    168         for(i=0;i<=k-1;i++)
    169         {
    170             if(c[i]=='u')printf("d");
    171             if(c[i]=='d')printf("u");
    172             if(c[i]=='l')printf("r");
    173             if(c[i]=='r')printf("l");
    174         }
    175         printf("
    ");
    176     }
    177     return 0;
    178 }
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  • 原文地址:https://www.cnblogs.com/tom987690183/p/3649082.html
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