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
ullddrurdllurdruldrSource
South Central USA 1998
题解
这题必须先把题意看懂，我被这个坑了…
这其实就是一个八数码问题，只不过我用的是\(BFS+A*\)，这题的\(A*\)本质就是一个运用贪心的剪枝
我们将每一个数到它的目标位置的曼哈顿距离的和定为该状态的估价值
由于可能会出现循环，所以我们必须判重；对于这题，\(Hash\)是一个很好的选择
至于\(Hash\)，我们用康托展开
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<bitset>
#include<iomanip>
#include<set>
#include<map>
using namespace std;
struct node
{
int f,g,id[10],mp[10];
string s;
bool operator< (const node &tmp) const
{
return f>tmp.f;
}
};
node q;
priority_queue<node> Q;
int d[10][10],goal[10];
int fac[9]={1,2,6,24,120,720,5040,40320,362880};
bool Hash[362881];
void Init()
{
for(int i=1;i<=3;++i)
for(int j=1;j<=3;++j)
for(int x=1;x<=3;++x)
for(int y=1;y<=3;++y)
d[(i-1)*3+j][(x-1)*3+y]=abs(i-x)+abs(j-y);
goal[0]=9,goal[1]=1,goal[2]=2,
goal[3]=3,goal[4]=4,goal[5]=5,
goal[6]=6,goal[7]=7,goal[8]=8;
}
void Read()
{
char c;
for(int i=1;i<=9;++i)
{
c=getchar();
while(c==' ') c=getchar();
if(c=='x') q.id[0]=i,q.mp[i]=0;
else q.id[(c^48)]=i,q.mp[i]=(c^48);
}
}
bool Check()
{
for(int i=0;i<=8;++i)
if(goal[i]!=q.id[i]) return 0;
return 1;
}
int A_star(node nx)
{
int Res=0;
for(int i=0;i<=8;++i) Res+=d[nx.id[i]][goal[i]];
return Res;
}
int Calc(node nx)
{
int Res=0,c;
for(int i=0;i<8;++i)
{
c=0;
for(int j=i+1;j<=8;++j)
if(nx.id[i]>nx.id[j]) ++c;
Res+=c*fac[i];
}
return Res;
}
void BFS()
{
Q.push(q);
Hash[Calc(q)]=1;
node nx;
int YouHua;
while(!Q.empty())
{
q=Q.top(); Q.pop();
if(Check())
{
cout<<q.s<<endl;
return;
}
if(q.id[0]<7)
{
nx=q,
nx.g=q.g+1,
nx.mp[q.id[0]]=q.mp[q.id[0]+3],
nx.mp[q.id[0]+3]=0,
nx.id[0]=q.id[0]+3,
nx.id[q.mp[q.id[0]+3]]=q.id[0],
YouHua=Calc(nx);
if(!Hash[YouHua])
{
Hash[YouHua]=1,
nx.s=q.s+'d',
nx.f=nx.g+A_star(nx);
Q.push(nx);
}
}
if(q.id[0]%3!=1)
{
nx=q,
nx.g=q.g+1,
nx.mp[q.id[0]]=q.mp[q.id[0]-1],
nx.mp[q.id[0]-1]=0,
nx.id[0]=q.id[0]-1,
nx.id[q.mp[q.id[0]-1]]=q.id[0],
YouHua=Calc(nx);
if(!Hash[YouHua])
{
Hash[YouHua]=1,
nx.s=q.s+'l',
nx.f=nx.g+A_star(nx);
Q.push(nx);
}
}
if(q.id[0]%3)
{
nx=q,
nx.g=q.g+1,
nx.mp[q.id[0]]=q.mp[q.id[0]+1],
nx.mp[q.id[0]+1]=0,
nx.id[0]=q.id[0]+1,
nx.id[q.mp[q.id[0]+1]]=q.id[0],
YouHua=Calc(nx);
if(!Hash[YouHua])
{
Hash[YouHua]=1,
nx.s=q.s+'r',
nx.f=nx.g+A_star(nx);
Q.push(nx);
}
}
if(q.id[0]>3)
{
nx=q,
nx.g=q.g+1,
nx.mp[q.id[0]]=q.mp[q.id[0]-3],
nx.mp[q.id[0]-3]=0,
nx.id[0]=q.id[0]-3,
nx.id[q.mp[q.id[0]-3]]=q.id[0],
YouHua=Calc(nx);
if(!Hash[YouHua])
{
Hash[YouHua]=1,
nx.s=q.s+'u',
nx.f=nx.g+A_star(nx);
Q.push(nx);
}
}
}
}
int main()
{
Init(),Read(),BFS();
return 0;
}