Submission #902508

#TimeUsernameProblemLanguageResultExecution timeMemory
902508Tuanlinh123Chorus (JOI23_chorus)C++17
87 / 100
7084 ms818372 KiB
#pragma GCC optimize("O3,unroll-loops") #pragma GCC target("avx2,fma,bmi,bmi2,popcnt,lzcnt") #include<bits/stdc++.h> #define ll long long #define pll pair<ll, ll> #define pb push_back #define mp make_pair #define fi first #define se second #define ld long double using namespace std; const long long inf=1e6, maxn=1e6+5; ll cnt[maxn], to[maxn]; long long n, k, dp[maxn], pre[maxn]; struct Line { long long a=0, b=inf*inf; ll idx=0; Line(){} Line(long long a, long long b, ll idx): a(a), b(b), idx(idx){} inline pair<long long, ll> ask(ll x){return {a*x+b, idx};} }; struct Node { Line A; ll l=0, r=0; }; Node LC[maxn*20]; struct Lichao { ll crr; Lichao() { crr=0; LC[0]=Node(); } inline void addline(Line A) { ll Start=-inf, End=inf, i=0; while (1) { ll mid=(Start+End)/2; bool l=A.ask(Start).fi<LC[i].A.ask(Start).fi; bool m=A.ask(mid).fi<LC[i].A.ask(mid).fi; if (m) swap(A, LC[i].A); if (A.a==inf) break; if (Start+1!=End) { if (l!=m) { if (!LC[i].l) { LC[i].l=++crr; LC[crr]=Node(); } i=LC[i].l; End=mid; } else { if (!LC[i].r) { LC[i].r=++crr; LC[crr]=Node(); } i=LC[i].r; Start=mid; } } else break; } } inline pair<long long, ll> query(ll x) { ll Start=-inf, End=inf, i=0; pair<long long, ll> ans={inf*inf, 0}; while (1) { ll mid=(Start+End)/2; ans=min(ans, LC[i].A.ask(x)); if (x<mid && LC[i].l) { i=LC[i].l; End=mid; } else if (x>=mid && LC[i].r) { i=LC[i].r; Start=mid; } else break; } return ans; } }; pair<long long, ll> solve(long long C) { ll crr=1; Lichao A; A.addline(Line(0, -pre[to[0]-1], 0)); for (ll i=1; i<=n; i++) dp[i]=inf*inf, cnt[i]=0; for (ll i=1; i<=n; i++) { while (crr<i && to[crr]<=i) A.addline(Line(-crr, -(pre[to[crr]-1]-(to[crr]-1)*crr)+dp[crr], cnt[crr])), crr++; tie(dp[i], cnt[i])=A.query(i); dp[i]+=C+pre[i], cnt[i]++; if (crr<i && dp[crr]+C<dp[i]) dp[i]=dp[crr]+C, cnt[i]=cnt[crr]+1; } return {dp[n], cnt[n]}; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); cin >> n >> k; string s; cin >> s; long long x=0, y=0, crr=1; for (ll i=0; i<n*2; i++) { if (s[i]=='A') x++, pre[x]=pre[x-1]+y; else y++; } for (ll i=1; i<=n; i++) while (crr<=pre[i]-pre[i-1]) to[crr]=i, crr++; while (crr<=n) to[crr]=n+1, crr++; for (ll i=0; i<=n; i++) if (to[i]<=i) to[i]=i+1; long long lo=0, hi=1LL*n*n; while (hi>lo) { long long mid=(lo+hi)/2; if (solve(mid).se<=k) hi=mid; else lo=mid+1; } cout << 0LL+solve(lo).fi-lo*k; }
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