标签:ret ++ space typedef lap zoj lin encode ext
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;
}
标签:ret ++ space typedef lap zoj lin encode ext
原文地址:http://www.cnblogs.com/clrs97/p/6321773.html