제출 #573265

#제출 시각아이디문제언어결과실행 시간메모리
573265MohamedAliSaidaneBoarding Passes (BOI22_passes)C++14
55 / 100
1511 ms4440 KiB
#include<bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> using namespace __gnu_pbds; using namespace std; typedef tree<int,null_type,less<int>,rb_tree_tag, tree_order_statistics_node_update> indexed_set; typedef long long ll; typedef long double ld; typedef pair<int,int> pii; typedef pair<ll,ll> pll; typedef pair<ld,ld> pld; typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pii> vpi; typedef vector<pll> vpl; #define pb push_back #define popb pop_back #define pp pop_back #define pf push_front #define popf pop_front #define all(x) (x).begin(),(x).end() #define ff first #define ss second ///#define int ll int nx[4] = {0,0,1,-1}, ny[4] = {1,-1,0,0}; ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} ll lcm(ll a , ll b) {return (a * b) / gcd(a , b);} const int MOD = 1e9 + 7; const int nax = 1e5 + 4; int n, g, sz = 1; ld dp[(1 << 15)][15]; string s; vi A[15]; vi st; ld exp_front[nax], exp_back[nax]; void update(int i, int val) { i += sz; st[i] = val; i /= 2; while(i) { st[i] = st[2 * i] + st[2 * i + 1]; i /= 2; } } int query(int p, int l, int r, int i, int j) { if( i> j) return 0; if(l >= i && r <= j) return st[p]; int m = (l+r)/2; return query(2 * p,l,m,i,min(j,m)) + query(2 * p + 1,m+1,r,max(i,m+1),j); } ld f(int mask, int b) { if(fabs(dp[mask][b] + 1) > 1e-9) return dp[mask][b]; ld rep = 0, ans = MOD; for(auto e: A[b]) { ld addfront = (ld)(query(1,0,sz-1,0,e)); ld addback = (ld)(query(1,0,sz - 1,e,n-1)); rep += min(exp_front[e] + addfront,exp_back[e] + addback); } for(auto e: A[b]) update(e,1); if(mask == (1 << g)-1) ans = 0; for(int j = 0; j < g; j++) { if((1 << j) & mask) continue; ans = min(ans,f(mask + (1 << j),j)); } for(auto e: A[b]) update(e,0); //cout << mask << ' ' << b << ' ' << ans << ' ' << rep << '\n'; return dp[mask][b] = ans + rep; } void solve() { cin >> s; n = s.length(); while(sz < n) sz = (sz << 1); st.assign(2 * sz + 1,0); for(int i = 0; i < n ;i ++) { int c =s[i] - 'A'; A[c].pb(i); g= max(g,c + 1); } for(int i = 0; i < (1 << g); i++) { for(int j = 0; j < g; j++) dp[i][j] =-1; } for(int group = 0; group < g; group ++) { int szi = A[group].size(); for(int i =0; i < szi; i++) { exp_front[A[group][i]] = (ld)((i * (i + 1))/2); exp_front[A[group][i]] /= (ld)(i+1); exp_back[A[group][i]] = (ld)(((szi -i - 1)*(szi - i))/2); exp_back[A[group][i]] /= (ld)(szi - i); //cout << A[group][i] << ' ' << exp_front[A[group][i]] << ' ' << exp_back[A[group][i]] << '\n'; } } ld ans = MOD; for(int i = 0; i < g; i++) ans = min(ans,f((1 << i),i)); cout << setprecision(40) << ans; } int32_t main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); int tt = 1; while(tt --) solve(); }
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