Eight
| Time Limit: 1000MS | Memory Limit: 65536K | |||
| Total Submissions: 41040 | Accepted: 16901 | 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
题目大意与分析
输入一个3x3的棋盘，每次将0上下左右移动，直到变换成为123456780的状态，求最短路线
BFS+Contor变换，康托展开公式：
以此求得的是在当前状态前面的状态数量
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <queue>
#include <string>
using namespace std;
struct node
{
int s[9];
string road;
int order;
int add0;
};
struct node start;
int jiecheng[10],f[2][4]={0,0,-1,1,-1,1,0,0},vis[400005],flag=0;
string anss;
char fang[5]="lrud";
int Contor(int s[],int n)
{
int order=0;
for(int i=0;i<n;i++)
{
int num=0;
for(int j=i+1;j<n;j++)
{
if(s[j]<s[i])
num++;
}
order+=num*jiecheng[n-i-1];
}
return order+1;
}
void bfs()
{
queue<struct node>q;
q.push(start);
vis[start.order]=1;
while(!q.empty())
{
struct node now=q.front();
if(now.order==46234)
{
flag=1;
anss=now.road;
break;
}
q.pop();
int x=now.add0/3;
int y=now.add0%3;
int i;
for(i=0;i<4;i++)
{
int xx=x+f[0][i];
int yy=y+f[1][i];
if(xx>=0&&xx<3&&yy>=0&&yy<3)
{
int add1=xx*3+yy;
struct node next=now;
next.add0=add1;
next.s[now.add0]=next.s[add1];
next.s[add1]=0;
next.add0=add1;
next.order=Contor(next.s,9);
if(vis[next.order]==0)
{
vis[next.order]=1;
next.road=now.road+fang[i];
q.push(next);
}
}
}
}
}
int main()
{
char x;
for(int i=0;i<9;i++)
{
cin>>x;
if(x==‘x‘)
{
start.s[i]=0;
start.add0=i;
}
else
start.s[i]=x-‘0‘;
}
start.order=Contor(start.s,9);
jiecheng[0]=1;
for(int i=1;i<=9;i++)
{
jiecheng[i]=i*jiecheng[i-1];
}
bfs();
if(flag)
cout<<anss<<endl;
else
cout<<"unsolvable"<<endl;
}
原文地址：https://www.cnblogs.com/dyhaohaoxuexi/p/12655476.html