Eight
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 30176 | Accepted: 13119 | Special Judge | ||
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
题目链接：POJ 1077
主要就是用康托展开来映射判重的问题，info::val就是康托展开hash值，info::step就是积累的状态。另外感觉这题剧毒，自己本来用int vis[]和char his[]想最后回溯记录答案从而代替速度比较慢的string，结果居然超时……TLE一晚上，要不是看了大牛的博客估计要一直T在这个坑点上。还有不知道为什么string的加号重载在C++编译器里会CE，换G++才过。
什么是康托展开？——康托展开介绍文章
代码：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=362880+20;
int fact[10]={1,1,2,6,24,120,720,5040,40320,362880};
int direct[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
char MOVE[5]="udlr";
struct info
{
int s[9];
int indx;
string step;
int val;
};
info S,E;
int T;
int vis[N];
string ans;
int calcantor(int s[])
{
int r=0;
for (int i=0; i<9; ++i)
{
int k=0;
for (int j=i+1; j<9; ++j)
{
if(s[j]<s[i])
++k;
}
r=r+k*fact[8-i];
}
return r;
}
bool check(const int &x,const int &y)
{
return (x>=0&&x<3&&y>=0&&y<3);
}
bool bfs()
{
CLR(vis,0);
queue<info>Q;
Q.push(S);
vis[S.val]=1;
info now,v;
while (!Q.empty())
{
now=Q.front();
Q.pop();
if(now.val==T)
{
ans=now.step;
return true;
}
for (int i=0; i<4; ++i)
{
int x=now.indx/3;
int y=now.indx%3;
x+=direct[i][0];
y+=direct[i][1];
if(check(x,y))
{
v=now;
v.indx=x*3+y;
v.s[now.indx]=v.s[v.indx];
v.s[v.indx]=0;
v.val=calcantor(v.s);
if(!vis[v.val])
{
vis[v.val]=1;
v.step=now.step+MOVE[i];
if(v.val==T)
{
ans=v.step;
return true;
}
Q.push(v);
}
}
}
}
return false;
}
int main(void)
{
char temp;
int i;
T=46233;
while (cin>>temp)
{
if(temp==‘x‘)
{
S.s[0]=0;
S.indx=0;
}
else
S.s[0]=temp-‘0‘;
for (i=1; i<9; ++i)
{
cin>>temp;
if(temp==‘x‘)
{
S.s[i]=0;
S.indx=i;
}
else
S.s[i]=temp-‘0‘;
}
S.val=calcantor(S.s);
puts(!bfs()?"unsolvable":ans.c_str());
}
return 0;
}