hdu 1043 Eight

我不是女神ヾ 2021-09-29 06:08 233阅读 0赞

# Eight #

**Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 25454    Accepted Submission(s): 6781  
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) ][South Central USA 1998 _Sepcial Judge Module By JGShining]

bfs+优先队列

![ContractedBlock.gif][] ![ExpandedBlockStart.gif][]

1 # include<string.h>
      2 # include<queue>
      3 # define N 363000
      4 using namespace std;
      5 bool visit[N];
      6 char visit1[N];
      7 int pre[N],st,a[10],ed;
      8 int dir[9]={
          1,1,2,6,24,120,720,5040,40320};
      9 struct node{
     10     int ma[10];
     11     int ans1;
     12     int x;
     13     int f;
     14     int g;
     15     bool operator <(const node &a)const {
     16         return a.f < f;//优先访问f较小者
     17     }
     18 };
     19 int hsh(int s[])
     20 {
     21     int i,j,cnt,sum;
     22     sum=0;
     23     for(i=1;i<=9;i++)
     24     {
     25         cnt=0;
     26         for(j=1;j<i;j++)
     27             if(s[j]>s[i]) cnt++;
     28             sum+=cnt*dir[i-1];
     29     }
     30     return sum;
     31 }
     32 int ABS(int x) {
          return x<0?(-x):x;}
     33 int h(int s[])//不算x时的曼哈顿距离
     34 {
     35     int curx,cury,endx,endy,sum,i,ans;
     36     sum=0;
     37     for(i=1;i<=9;i++)
     38     {
     39         if(s[i]==9) continue;
     40         ans=s[i];
     41         curx=(i+2)/3;
     42         cury=(i-1)%3+1;
     43         endx=(ans+2)/3;
     44         endy=(ans-1)%3+1;
     45         sum=sum+ABS(curx-endx)+ABS(cury-endy);
     46     }
     47     return sum;
     48 }
     49 void bfs()
     50 {
     51     int ans,i;
     52     priority_queue<node>q;
     53     node cur,next;
     54     cur.ans1=st=hsh(a);
     55     visit[cur.ans1]=1;
     56     if(st==ed) return;
     57     for(i=1;i<=9;i++)
     58     {
     59         cur.ma[i]=a[i];
     60         if(a[i]==9) cur.x=i;
     61     }
     62     cur.g=0;//表示深度
     63     cur.f=h(a);
     64     q.push(cur);
     65     while(!q.empty())
     66     {
     67         cur=q.top();
     68         q.pop();
     69         if((cur.x+2)/3!=1) //向上翻
     70         {
     71             next=cur;
     72             next.x=cur.x-3;
     73             next.ma[cur.x]=next.ma[next.x];
     74             next.ma[next.x]=9;
     75             ans=hsh(next.ma);
     76             if(!visit[ans])
     77             {
     78                 next.g++;
     79                 next.f=next.g+h(next.ma);
     80                 visit[ans]=1;
     81                 next.ans1=ans;
     82                 pre[ans]=cur.ans1;
     83                 visit1[ans]='u';
     84                 if(ans==ed) return;
     85                 q.push(next);
     86             }
     87         }
     88         if((cur.x+2)/3!=3)//向下翻
     89         {
     90             next=cur;
     91             next.x=cur.x+3;
     92             next.ma[cur.x]=next.ma[next.x];
     93             next.ma[next.x]=9;
     94             ans=hsh(next.ma);
     95             if(!visit[ans])
     96             {
     97                 next.g++;
     98                 next.f=next.g+h(next.ma);
     99                 visit[ans]=1;
    100                 next.ans1=ans;
    101                 pre[ans]=cur.ans1;
    102                 visit1[ans]='d';
    103                 if(ans==ed) return;
    104                 q.push(next);
    105             }
    106         }
    107         if(cur.x%3!=1)//向左翻
    108         {
    109             next=cur;
    110             next.x=cur.x-1;
    111             next.ma[cur.x]=next.ma[next.x];
    112             next.ma[next.x]=9;
    113             ans=hsh(next.ma);
    114             if(!visit[ans])
    115             {
    116                 next.g++;
    117                 next.f=next.g+h(next.ma);
    118                 visit[ans]=1;
    119                 next.ans1=ans;
    120                 pre[ans]=cur.ans1;
    121                 visit1[ans]='l';
    122                 if(ans==ed) return;
    123                 q.push(next);
    124             }
    125         }
    126         if(cur.x%3!=0)//向右翻
    127         {
    128             next=cur;
    129             next.x=cur.x+1;
    130             next.ma[cur.x]=next.ma[next.x];
    131             next.ma[next.x]=9;
    132             ans=hsh(next.ma);
    133             if(!visit[ans])
    134             {
    135                 next.g++;
    136                 next.f=next.g+h(next.ma);
    137                 visit[ans]=1;
    138                 next.ans1=ans;
    139                 pre[ans]=cur.ans1;
    140                 visit1[ans]='r';
    141                 if(ans==ed) return;
    142                 q.push(next);
    143             }
    144         }
    145     }
    146 }
    147 int check(int s[])
    148 {
    149     int i,j,cnt=0;
    150     for(i=1;i<=9;i++)
    151     {
    152         if(s[i]==9) continue;
    153         for(j=1;j<i;j++)
    154         {
    155             if(s[j]==9) continue;
    156             if(s[j]>s[i]) cnt++;
    157         }
    158     }
    159     return cnt;
    160 }
    161 int main()
    162 {
    163     int i,j,ans;
    164     char str[50];
    165     while(gets(str))
    166     {
    167         ans=0;
    168         memset(visit,0,sizeof(visit));
    169         for(i=0;str[i];i++)
    170             if(str[i]=='x') a[++ans]=9;
    171             else if(str[i]!=' ') a[++ans]=str[i]-'0';
    172             ed=0;
    173         ans=check(a);
    174         if(ans%2) {puts("unsolvable");continue;}
    175             bfs();
    176             j=0;
    177             while(ed!=st)
    178             {
    179                 str[j++]=visit1[ed];
    180                 ed=pre[ed];
    181             }
    182             for(i=j-1;i>=0;i--)
    183                 printf("%c",str[i]);
    184                 puts("");
    185     }
    186     return 0;
    187 }

转载于:https://www.cnblogs.com/52why/p/7483099.html

[South Central USA 1998 _Sepcial Judge Module By JGShining]: http://acm.split.hdu.edu.cn/search.php?field=problem&key=South+Central+USA+1998+%28Sepcial+Judge+Module+By+JGShining%29&source=1&searchmode=source
[ContractedBlock.gif]: https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif
[ExpandedBlockStart.gif]: /images/20210726/4348202a89dc4811ae9e537c61d6c60f.png