BZOJ1330 : Editing a Book

注意到答案不超过$5$，因此可以考虑BFS求出距离起始态或者终止态不超过$2$的所有状态。

设它们到起始态、终止态的距离分别为$f[S],g[S]$，则$ans=\min(5,f[S]+g[S])$。

时间复杂度$O(n^6\log(n!))$。

 

#include<cstdio>
#include<algorithm>
using namespace std;
typedef unsigned long long ll;
const int N=12,M=363000;
int n,i,j,k,a[N],len[N],S,x,z,v[M],g[M],h,t,q[M],ans;ll f[N][M];
inline ll encode(){
  ll t=0;
  for(int i=0;i<n;i++)t=t<<4|a[i];
  return t;
}
inline int getid(ll x){
  int l=0,r=len[n]-1;
  while(l<=r){
    int mid=(l+r)>>1;
    if(f[n][mid]==x)return mid;
    if(f[n][mid]<x)l=mid+1;else r=mid-1;
  }
}
inline void ext(int x){
  if(~v[x])return;
  v[q[++t]=x]=z;
}
int main(){
  for(n=2;n<=9;n++){
    for(i=0;i<n;i++)a[i]=i;
    do{f[n][len[n]++]=encode();}while(next_permutation(a,a+n));
    sort(f[n],f[n]+len[n]);
  }
  while(~scanf("%d",&n)){
    if(!n)return 0;
    for(i=0;i<n;i++)scanf("%d",&a[i]),a[i]--;
    if(n==1){puts("0");continue;}
    S=getid(encode());
    h=1,t=0;
    for(i=0;i<len[n];i++)v[i]=-1;
    z=0;
    ext(S);
    while(h<=t){
      z=v[x=q[h++]]+1;
      if(z>2)continue;
      ll O=f[n][x];
      for(i=1;i<n;i++)for(j=i;j<n;j++){
        ll U=(1ULL<<((j-i+2)*4))-1,F=O;
        int L=(j-i+1)*4;
        for(k=4*(i-1);k>=0;k-=4){
          ll A=(F>>k)&U;
          F^=(A^((A>>4)|(A&15)<<L))<<k;
          ext(getid(F));
        }
      }
    }
    for(i=0;i<len[n];i++)g[i]=v[i],v[i]=-1;
    h=1,t=z=0;
    ext(0);
    while(h<=t){
      z=v[x=q[h++]]+1;
      if(z>2)continue;
      ll O=f[n][x];
      for(i=1;i<n;i++)for(j=i;j<n;j++){
        ll U=(1ULL<<((j-i+2)*4))-1,F=O;
        int L=(j-i+1)*4;
        for(k=4*(i-1);k>=0;k-=4){
          ll A=(F>>k)&U;
          F^=(A^((A>>4)|(A&15)<<L))<<k;
          ext(getid(F));
        }
      }
    }
    for(ans=5,i=0;i<len[n];i++)if(~v[i]&&~g[i])ans=min(ans,v[i]+g[i]);
    printf("%d\n",ans);
  }
  return 0;
}