Eight HDU - 1043
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.
题意:
由15个滑块构造的,每个滑块(瓷砖由1-15组成)。x表示空格,r,l,d,u表示移动x的移动操作。题目让解决8个滑块构成的方格(个人觉得就是把蒙娜丽莎的微笑谜图还原)
输入:八个字符代表一个排列,把移动成12345678X
输出:如果可以找到解(12345678X)就输出操作,不然就输出“unsolvable”,注意输出的格式。
解题思路:还是和前一个博客一样的思路,首先打表–1.编码 2康托展开 3.记录父节点和操作字符 4.判断移动的合法性
#include
#include
#include
#include
#include
using namespace std;
const int maxn=370000;
int kangtuo[10];
char op[maxn];//记录到来的操作符
int fa[maxn];//记录父节点
int vis[maxn];//记录是否可以到达的状态
char dir[]="urld";//注意因为是打表,反向的所以 上下左右是反的 !!!
int dx[]={3,-1,1,-3};
int get_code(string s)
{
int len=s.size();
int ans=0;
for(int i=0;iint cnt=0;
for(int j=i+1;jif(s[i]-'0'>s[j]-'0') cnt++;
ans+=cnt*kangtuo[8-i];
}
// cout<<"get_code: "<
return ans;
}
int move(int w,int x)
{
if(dx[x]==1&&w%3==2) return 0;
if(dx[x]==-1&&w%3==0) return 0;
if(dx[x]==-3&&w<3) return 0;
if(dx[x]==3&&w>5) return 0;
return 1;
}
void bfs(string s,int u)
{
memset(vis,0,sizeof(vis));
int tmp=get_code(s);
vis[tmp]=1;
op[tmp]='0';
fa[tmp]=tmp;
queue<string> q;
queue<int> wei;
q.push(s);
wei.push(u);
while(!q.empty())
{
string str=q.front();
int w=wei.front();
q.pop(),wei.pop();
int str_code=get_code(str);
for(int i=0;i<4;i++)
{
int nx=w+dx[i];
if(move(w,i))
{
string str2=str;
str2[nx]=str[w];
str2[w]=str[nx];
// cout<
int tt=get_code(str2);
if(!vis[tt])
{
vis[tt]=1;
op[tt]=dir[i];//dlru
fa[tt]=str_code;//前一个
q.push(str2);
wei.push(nx);
}
}
}
}
}
int main()
{
kangtuo[0]=1;
for(int i=1;i<9;i++)
kangtuo[i]=kangtuo[i-1]*i;//cout<
string str="123456780";
bfs(str,8);//打表
int code_str=get_code(str);
string s;
while(getline(cin,s))
{
for(int i=0;iif (s[i]==' ') s.erase(i,1),i--;
if(s[i]=='x') s[i]='0';
}
//cout<
int code_s=get_code(s);
//cout<
if(!vis[code_s])
printf("unsolvable\n") ;
else
{
while(code_s!=code_str)
{
printf("%c",op[code_s]);
code_s=fa[code_s];
//cout<
}
printf("\n");
}
}
return 0;
}