[kuangbin]专题二 搜索进阶 Eight HDU - 1043【反向BFS】【康托展开】【哈希】
作者：互联网
【题目描述】
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.
这个15字谜已经存在了100多年；即使你不知道这个名字，你也知道它。它是用15个滑动瓷砖建造的，每个瓷砖上的数字从1到15不等，所有这些都被包装成一个4×4的框架，一个瓷砖缺失了。让我们将丢失的瓷砖称为“x”；拼图的目标是排列这些瓷砖，以便按如下顺序排列：
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
唯一合法的操作是将“x”与其共享边缘的瓷砖之一交换。作为一个例子，下面的移动顺序解决了一个稍微混乱的谜题：
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->
字母表示x和哪个相邻的数字交换，合法的操作分别为“r”、“l”、“u”和“d”，分别表示右、左、上和下。
并不是所有的谜题都能解决；1870年，一个名叫山姆·洛伊德的人因散发无法解决的谜题而出名，并使许多人感到沮丧。实际上，要将一个常规的谜题变成一个无法解决的难题，所要做的就是交换两个瓷砖(当然，不包括丢失的‘x’瓷砖)。在这个问题中，你将编写一个程序来解决不太知名的8-谜题，由三次排列的瓷砖组成。
【输入】
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
【输出】
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.
【样例输入】
2 3 4 1 5 x 7 6 8
【样例输出】
ullddrurdllurdruldr
题目链接：https://cn.vjudge.net/problem/HDU-1043
写了两天看了无数大佬的代码终于过了QAQ，还是太菜了
临时学的康托展开，不会看这里 → https://www.cnblogs.com/sky-stars/p/11216035.html
照着这个写的 → https://www.cnblogs.com/sky-stars/p/11216007.html
代码如下：
#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
int mp[10];
static const int MAXN=362880;
static const int result=46234;
static const int fac[]={1,1,2,6,24,120,720,5040,40320,362880};
static const int dx[]={0,0,1,-1},dy[]={1,-1,0,0};
static string op="lrud"; //要反着写
bool vis[MAXN+10];
string path[MAXN+10];
struct Node{
string path;
int hash;
int pos;
};
void decantor(int hashs)
{
vector<int> v;
for(int i=0;i<=8;i++)
v.push_back(i);
for(int i=0;i<=8;i++)
{
int r=hashs%fac[8-i];
int t=hashs/fac[8-i];
hashs=r;
sort(v.begin(),v.end());
mp[i]=v[t];
v.erase(v.begin()+t);
}
}
int cantor()
{
int sum=0;
for(int i=0;i<=8;i++)
{
int cnt=0;
for(int j=i+1;j<=8;j++)
{
if(mp[j]<mp[i])
cnt++;
}
sum+=fac[8-i]*cnt;
}
return sum+1;
}
void bfs()//反向BFS，先把所有可能到达情况存下来
{
queue<Node> Q;
memset(vis,false,sizeof(vis));
for(int i=0;i<8;i++)
mp[i]=i+1;
mp[8]=0;
vis[result]=true;
path[result]="";
Q.push(Node{"",result,8});
while(!Q.empty())
{
Node now=Q.front();
Q.pop();
for(int i=0;i<4;i++)
{
Node next;
if(0<=now.pos/3+dx[i] && now.pos/3+dx[i]<=2 && 0<=now.pos%3+dy[i] && now.pos%3+dy[i]<=2)
{
next=now;
next.pos=(now.pos/3+dx[i])*3+now.pos%3+dy[i];
decantor(now.hash-1);
swap(mp[now.pos],mp[next.pos]);
next.hash=cantor();
if(!vis[next.hash])
{
vis[next.hash]=true;
next.path=op[i]+next.path;
path[next.hash]=next.path;
Q.push(next);
}
}
}
}
}
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(0),cout.tie(0);
bfs();
char ch;
while(cin>>ch)//多组数据输入
{
if(ch=='x')
mp[0]=0;
else
mp[0]=ch-'0';
for(int i=1;i<=8;i++)
{
cin>>ch;
if(ch=='x')
mp[i]=0;
else
mp[i]=ch-'0';
}
int hashs=cantor();
if(!vis[hashs])
cout<<"unsolvable"<<endl;//vis没有被改动则说明无法到达此情况
else
cout<<path[hashs]<<endl;
}
return 0;
}
标签：1043,10,HDU,12,15,int,13,11,康托 来源： https://blog.csdn.net/inv00ker/article/details/98470437