Submission #598798

#	Time	Username	Problem	Language	Result	Execution time	Memory
598798		zaneyu	Boarding Passes (BOI22_passes)	C++14	0 / 100	38 ms	1100 KiB

passes

/*input
HEHEHEHIHILOL
*/
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long,null_type,less_equal<long long>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
//#pragma GCC target("avx2")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<ll,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET(v,x) lower_bound(ALL(v),x)-v.begin()
const ll maxn=1e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
    out<<P.f<<' '<<P.s;
    return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
    REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?"\n":"\n");
    return out;
}
ll mult(ll a,ll b){
    return a*b%MOD;
}
ll mypow(ll a,ll b){
    a%=MOD;
    if(a==0) return 0;
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
const int cc=15;
ld dp[(1<<cc)];
ll pf[20][maxn],sf[20][maxn];
vector<int> v[20];
ld get(int m,int i,ld p){
    ll tot=0;
    REP(j,cc){
        if(m&(1<<j)){
            tot+=pf[j][v[i][p]]+sf[j][v[i][p+1]];
        }
    }
    ld k=sz(v[i])-p-2;
    return tot+p*(p-1)/4+k*(k-1)/4;
}
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    string s;
    cin>>s;
    int n=sz(s);
    REP(i,cc) v[i].pb(0);
    REP(i,n) v[s[i]-'A'].pb(i+1);
    REP(i,cc) v[i].pb(n+1);
    REP(i,cc){
        REP1(j,n){
            pf[i][j]=lower_bound(ALL(v[i]),j)-v[i].begin()-1;
            sf[i][j]=sz(v[i])-pf[i][j]-2;
        }
    }
    REP(k,cc){
        REP(i,cc){
            REP1(j,sz(v[i])-1){
                pf[k][v[i][j]]+=pf[k][v[i][j-1]];
            }
            for(int j=sz(v[i])-2;j>=0;j--){
                sf[k][v[i][j]]+=sf[k][v[i][j+1]];
            }
        }
    }
    REP1(i,(1<<cc)-1){
        dp[i]=INF64;
        REP(j,cc){
            if(i&(1<<j)){
                if(sz(v[j])==2){
                    MNTO(dp[i],dp[i^(1<<j)]);
                    continue;
                }
                int z=(i^(1<<j));
                int l=0,r=sz(v[j])-2;
                while(r-l>2){
                    int m1=(r-l)/3+l,m2=r-(r-l)/3;
                    if(get(z,j,m1)>get(z,j,m2)) l=m1+1;
                    else r=m2;
                }
                ld ans=INF64;
                for(int k=l;k<=r;k++) MNTO(ans,get(z,j,k));
                MNTO(dp[i],dp[i^(1<<j)]+ans);
            }
        }
    }
    cout<<fixed<<setprecision(20)<<dp[(1<<cc)-1];
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 25
• Subtask 3: 0 / 30
• Subtask 4: 0 / 40