ZOJ 3814 Sawtooth Puzzle status compression search

ZOJ 3814 Sawtooth Puzzle status compression search

Since there are only four statuses in a box, the total number of statuses is only 4 ^ 9, and bfs can be solved.

The trouble lies in simulation.

The method of saving my status is to regard the initial status as 000000000. If it is rotated once clockwise, + 1, 3 + 1 = 0.

During the bfs process, you need to set a dfs calculation to which boxes will be affected by rotating the current box.

Note that there are more than one target status, because some boxes remain unchanged after rotation, we must calculate all possible target statuses, the box in the middle of the sample is the case.

#include 
 
  #include
  
   #include#include
   
    #include
    
     #include
     
      #include
      #include
       
        using namespace std;#define maxn 100005int h[15];struct node{ int x,dis;}t,f;char st[10][10][10];char ed[10][10][10];char tst[10][10];char tzf[10][10];vector
        
          fin[15];int edge[10][4];struct fuck{ int to,turn; fuck(int a,int b){to=a;turn=b;} fuck(){}}v[20];int top;bool use[15];bool FINAL[1111111];void cal(int now){ memcpy(tst,st[now],sizeof(st[now])); for(int d=0;d<4;d++) { if(memcmp(tst,ed[now],sizeof(tst))==0) {fin[now].push_back(d);} for(int i=0;i<8;i++) { for(int j=0;j<8;j++) { tzf[j][8-i-1]=tst[i][j]; } } memcpy(tst,tzf,sizeof(tzf)); }}bool vis[1111111];int Turn[9];bool isok(int x,int y){ return x>=0&&x<3&&y>=0&&y<3;}int dx[]={0,-1,0,1};int dy[]={-1,0,1,0};int op(int a,int b){ if(a-b>=0) return a-b; return 3-(b-a-1);}bool can(int a,int b,int d){ if(d==0&&edge[a][op(0,Turn[a])]&&edge[b][op(2,Turn[b])]) return true; if(d==1&&edge[a][op(1,Turn[a])]&&edge[b][op(3,Turn[b])]) return true; if(d==2&&edge[a][op(2,Turn[a])]&&edge[b][op(0,Turn[b])]) return true; if(d==3&&edge[a][op(3,Turn[a])]&&edge[b][op(1,Turn[b])]) return true; return false;}void dfs(int x,int y,int flag){ v[top++]=fuck(x*3+y,flag); for(int d=0;d<4;d++) { int nx=x+dx[d]; int ny=y+dy[d]; if(isok(nx,ny)&&use[nx*3+ny]==0&&can(x*3+y,nx*3+ny,d)) { use[nx*3+ny]=true; dfs(nx,ny,-flag); } }}void bfs(){ memset(Turn,0,sizeof(Turn)); memset(vis,0,sizeof(vis)); vis[0]=true; queue
         
           q; f.dis=0;f.x=0; q.push(f); while(!q.empty()) { f=q.front();q.pop(); for(int i=0;i<9;i++) Turn[i]=(f.x/h[9-i-1])%4; for(int d=0;d<9;d++) { t=f; t.dis++; top=0; memset(use,0,sizeof(use)); use[d]=1; dfs(d/3,d%3,1); int tmp=t.x; for(int i=0;i