POJ-3131-Cubic Eight-Puzzle(双向BFS+哈希)
作者:互联网
Description
Let’s play a puzzle using eight cubes placed on a 3 × 3 board leaving one empty square.
Faces of cubes are painted with three colors. As a puzzle step, you can roll one of the cubes to a adjacent empty square. Your goal is to make the specified color pattern visible from above by a number of such steps.
The rules of this puzzle are as follows.
-
Coloring of Cubes: All the cubes area colored in the same way as shown in Figure 1. The opposite faces have the same color.
Figure 1: Coloring of a cube
-
Initial Board State: Eight cubes are placed on the 3 × 3 board leaving one empty square. All the cubes have the same orientation as shown in Figure 2. As shown in the figure, squares on the board are given x and y coordinates, (1, 1), (1, 2), …, and (3, 3). The position of the initially empty square may vary.
Figure 2: Initial board state
-
Rolling Cubes: At each step, we can choose one of the cubes adjacent to the empty square and roll it into the empty square, leaving the original position empty. Figure 3 shows an example.
Figure 3: Rolling a cube
-
Goal: The goal of this puzzle is to arrange the cubes so that their top faces form the specified color pattern by a number of cube rolling steps described above.
Your task is to write a program that finds the minimum number of steps required to make the specified color pattern from the given initial state.
Input
The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets is less than 16. Each dataset is formatted as follows.
x y F11 F21 F31 F12 F22 F32 F13 F23 F33
The first line contains two integers x and y separated by a space, indicating the position (x, y) of the initially empty square. The values of x and y are 1, 2, or 3.
The following three lines specify the color pattern to make. Each line contains three characters F1j, F2j, and F3j, separated by a space. Character Fij indicates the top color of the cube, if any, at the position (i, j) as follows:
B:
Blue,
W:
White,
R:
Red,
E:
the square is Empty.
There is exactly one ‘E
’ character in each dataset.
Output
For each dataset, output the minimum number of steps to achieve the goal, when the goal can be reached within 30 steps. Otherwise, output “-1
” for the dataset.
Sample Input
1 2 W W W E W W W W W 2 1 R B W R W W E W W 3 3 W B W B R E R B R 3 3 B W R B W R B E R 2 1 B B B B R B B R E 1 1 R R R W W W R R E 2 1 R R R B W B R R E 3 2 R R R W E W R R R 0 0
Sample Output
0 3 13 23 29 30 -1 -1
Source
Japan 2006思路:数据太大,先抽象出状态,再哈希一下,然后就是双向BFS。
#include <stdio.h>
#include <string.h>
#include <stack>
using namespace std;
#define INF 99999999
struct{
int step,state;
}t,que1[1000000],que2[1000000];
stack<int>stk;
bool vis1[5000007],vis2[5000007];
int top1,top2,bottom1,bottom2,mp[9],nxt[2][7]={{0,4,6,5,1,3,2},{0,5,3,2,6,1,4}},head[5000007],next[5000007],val[5000007],total;
int hashval(int x)
{
int t,i;
t=x%5000007;
for(i=head[t];i!=-1;i=next[i])
{
if(val[i]==x) return i;
}
val[total]=x;
next[total]=head[t];
head[t]=total;
total++;
return total-1;
}
void pushtarget(int cnt,int state)
{
if(cnt==-1)
{
vis2[hashval(state)]=1;
que2[bottom2].step=0;
que2[bottom2].state=state;
bottom2++;
return;
}
if(mp[cnt]%2)
{
pushtarget(cnt-1,state*7+mp[cnt]);
pushtarget(cnt-1,state*7+mp[cnt]+1);
}
else pushtarget(cnt-1,state*7);
}
int getstate()
{
int temp=0;
for(int i=8;i>=0;i--)
{
temp=temp*7+mp[i];
}
return temp;
}
void spread(int x)
{
int i=0,temp,cnt,old;
while(i<9)
{
if(x%7==0) cnt=i;
mp[i++]=x%7;
x/=7;
}
cnt+=1;
if(cnt>=0 && cnt<9 && cnt%3>0)
{
old=mp[cnt];
mp[cnt-1]=nxt[0][mp[cnt]];
mp[cnt]=0;
stk.push(getstate());
mp[cnt]=old;
mp[cnt-1]=0;
}
cnt-=1;
cnt-=1;
if(cnt>=0 && cnt<9 && cnt%3<2)
{
old=mp[cnt];
mp[cnt+1]=nxt[0][mp[cnt]];
mp[cnt]=0;
stk.push(getstate());
mp[cnt]=old;
mp[cnt+1]=0;
}
cnt+=1;
cnt-=3;
if(cnt>=0 && cnt<9)
{
old=mp[cnt];
mp[cnt+3]=nxt[1][mp[cnt]];
mp[cnt]=0;
stk.push(getstate());
mp[cnt]=old;
mp[cnt+3]=0;
}
cnt+=3;
cnt+=3;
if(cnt>=0 && cnt<9)
{
old=mp[cnt];
mp[cnt-3]=nxt[1][mp[cnt]];
mp[cnt]=0;
stk.push(getstate());
mp[cnt]=old;
mp[cnt-3]=0;
}
cnt-=3;
}
int bfs()
{
int step1=0,step2=0,ans=INF,temp;
for(step1=0;step1<=20;step1++)
{
while(top1<bottom1 && que1[top1].step==step1)
{
spread(que1[top1].state);
while(!stk.empty())
{
temp=stk.top();
stk.pop();
if(vis2[hashval(temp)])
{
return step1+step2+1;
}
if(!vis1[hashval(temp)])
{
vis1[hashval(temp)]=1;
que1[bottom1].state=temp;
que1[bottom1].step=que1[top1].step+1;
bottom1++;
}
}
top1++;
}
while(top2<bottom2 && que2[top2].step==step2 && step2<9)
{
spread(que2[top2].state);
while(!stk.empty())
{
temp=stk.top();
stk.pop();
if(vis1[hashval(temp)]) return step1+step2+2;
if(!vis2[hashval(temp)])
{
vis2[hashval(temp)]=1;
que2[bottom2].state=temp;
que2[bottom2].step=step2+1;
bottom2++;
}
}
top2++;
}
if(step2<9) step2++;
}
return -1;
}
int main()
{
int x,y,i,j,temp;
char ctemp;
while(~scanf("%d%d",&y,&x) && x)
{
x--;
y--;
top1=top2=bottom1=bottom2=0;
memset(vis1,0,sizeof vis1);
memset(vis2,0,sizeof vis2);
memset(head,-1,sizeof head);
total=0;
while(!stk.empty()) stk.pop();
for(i=0;i<9;i++)
{
ctemp=getchar();
if(ctemp=='\n' || ctemp==' ')
{
i--;
continue;
}
if(ctemp=='W')mp[i]=1;
else if(ctemp=='B') mp[i]=3;
else if(ctemp=='R') mp[i]=5;
else if(ctemp=='E') mp[i]=0;
}
pushtarget(8,0);
temp=0;
for(i=8;i>=0;i--)
{
if(i!=x*3+y) temp=temp*7+1;
else temp*=7;
}
if(vis2[hashval(temp)])
{
printf("0\n");
continue;
}
vis1[hashval(temp)]=1;
que1[bottom1].step=0;
que1[bottom1].state=temp;
bottom1++;
printf("%d\n",bfs());
}
}
标签:cnt,square,temp,Cubic,int,Puzzle,BFS,state,mp 来源: https://www.cnblogs.com/ldxsuanfa/p/10700620.html