Eight
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10351 Accepted Submission(s): 2755
Special Judge
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.
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
should include no spaces and start at the beginning of the line. Do not print a blank line between cases.
2 3 4 1 5 x 7 6 8
ullddrurdllurdruldr
题意：
就是还原八数码。输出操作。
思路：
利用康拓展开判重。通过康拓展开知一共有362879种状态。用逆序数判断可行解见八数码可行解。然后宽搜。
详细见代码：
#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-8;
const double PI=acos(-1.0);
const int maxn=370010;//最大为362879
const int num=9;
const int over=46233;
typedef __int64 ll;
int c[10],head,tail;
int vis[maxn];
int fac[15]={1,1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600};//存阶乘
int dx[4]={-1,1,-3,3};//四个方向
map<int,char> mpp;
struct yb
{
int mp[10];
int bp,pre,code,dre;//空格位置。上一状态的下标(记录路径),康拓展开,方向。
} q[maxn],tp,tt;
/***********树状数组求逆序数(规模比较小快不了多少)***********/
int lowbit(int x)
{
return x&-x;
}
void update(int x)
{
while(x<=num)
{
c[x]++;
x+=lowbit(x);
}
}
int getsum(int x)
{
int sum=0;
while(x>0)
{
sum+=c[x];
x-=lowbit(x);
}
return sum;
}
/***************康拓展开***************/
ll contor(int arr[])
{
int i,j,ct;
ll sum=0;
for(i=0;i<num;i++)
{
ct=0;
for(j=i+1;j<num;j++)
if(arr[j]<arr[i])
ct++;
sum+=ct*fac[num-i-1];
}
return sum;
}
bool ok(int px,int x,int dre)//判断是否越界
{
int lim=((px+3)/3)*3-1;
if(x<0||x>=num)
return false;
if(dre>1)
return true;
if(x>lim||x<lim-2)
return false;
return true;
}
void bfs()
{
int i,x;
memset(vis,0,sizeof vis);
vis[tp.code=contor(tp.mp)]=1;
tp.pre=-1;
head=tail=0;
q[tail++]=tp;
while(head<tail)
{
tp=q[head];
if(tp.code==over)
return;
for(i=0;i<4;i++)
{
x=tp.bp+dx[i];
if(ok(tp.bp,x,i))
{
tt=tp;
tt.mp[tt.bp]=tt.mp[x];
tt.mp[x]=0;
tt.code=contor(tt.mp);
if(!vis[tt.code])
{
vis[tt.code]=1;
tt.bp=x;
tt.pre=head;
tt.dre=i;
q[tail++]=tt;
}
}
}
head++;
}
}
void print(int x)//输出路径
{
if(q[x].pre==-1)
{
//show(q[x].mp);
return;
}
print(q[x].pre);
printf("%c",mpp[q[x].dre]);
//show(q[x].mp);
}
int main()
{
int i,ni,ct;
char com[10];
mpp.clear();
mpp[0]='l';//建立方向和下标的映射
mpp[1]='r';
mpp[2]='u';
mpp[3]='d';
while(~scanf("%s",com))
{
ct=0;
if(com[0]!='x')
tp.mp[0]=com[0]-'0';
else
tp.mp[0]=0,tp.bp=0;
for(i=1;i<9;i++)
{
scanf("%s",com);
if(com[0]!='x')
tp.mp[i]=com[0]-'0';
else
tp.mp[i]=0,tp.bp=i;
}
for(i=0;i<9;i++)
if(tp.mp[i])
tt.mp[ct++]=tp.mp[i];
ni=0;
memset(c,0,sizeof c);
for(i=0;i<8;i++)//求逆序数
{
update(tt.mp[i]);
ni+=i+1-getsum(tt.mp[i]);
}
if(ni&1)
{
printf("unsolvable\n");
continue;
}
bfs();
print(head);
printf("\n");
}
return 0;
}
/******************debug********************/
void show(int arr[])
{
int i,j;
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
{
printf("%d ",arr[i*3+j]);
}
printf("\n");
}
printf("--------------------\n");
getchar();
}