洛谷$P3226 [HNOI2012]$集合选数 状压$dp$

正解:$dp$

解题报告:

传送门$QwQ$

考虑列一个横坐标为比值为2的等比数列,纵坐标为比值为3的等比数列的表格.发现每个数要选就等价于它的上下左右不能选.

于是就是个状压$dp$板子了$QwQ$

然后因为有些数是无关联的就不会在一个表格中($eg:1,5$.所以要建多个表格,最后乘法原理就好,$over$

 

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define ri register int
#define rb register bool
#define rc register char
#define lowbit(x) (x&(-x))
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)

const int N=100000+10,mod=1000000001;
int n,as=1,cnt,num[35],f[35][N];//说下昂QwQ,就这里这个[N]不是原本的N的意义,,,只是恰好等于N.QwQ
bool vis[N];
vector<int>V[35];

il int read()
{
    rc ch=gc;ri x=0;rb y=1;
    while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
    if(ch=='-')ch=gc,y=0;
    while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
    return y?x:-x;
}
il void build(ri x)
{cnt=0;while(x<=n){ri tmp=x;num[++cnt]=0;while(tmp<=n)++num[cnt],vis[tmp]=1,tmp<<=1;x=1ll*x*3;}}
il bool check(ri x){ri pre=0;while(x){ri tmp=lowbit(x);if(tmp==(pre<<1))return 0;pre=tmp;x-=pre;}return 1;}
il void pre(){rp(i,1,cnt){V[i].clear();rp(j,0,(1<<num[i])-1)if(check(j))V[i].push_back(j);}}
il void inc(ri &x,ri y){x+=y;if(x>=mod)x-=mod;}
il int cal()
{
    pre();ri ret=0;
    rp(i,1,cnt){ri sz=V[i].size();rp(j,0,sz-1)f[i][j]=(i==1);}
    rp(i,2,cnt)
    {
        ri sznw=V[i].size(),szpr=V[i-1].size();
        rp(j,0,sznw-1){rp(k,0,szpr-1)if(!(V[i][j]&V[i-1][k]))inc(f[i][j],f[i-1][k]);if(i==cnt)inc(ret,f[i][j]);}
    }
    if(cnt==1)ret=V[1].size();
    return ret;
}

int main()
{
    //freopen("3226.in","r",stdin);freopen("3226.out","w",stdout);
    n=read();rp(i,1,n)if(!vis[i]){build(i);as=1ll*as*cal()%mod;}printf("%d
",as);
    return 0;
}