Submission #748664

#TimeUsernameProblemLanguageResultExecution timeMemory
748664inventiontimeBoarding Passes (BOI22_passes)C++17
5 / 100
5 ms1680 KiB
#include <bits/stdc++.h> using namespace std; #define int ll #define endl '\n' //comment for interactive #define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0) #define pb push_back #define re resize #define ff first #define ss second #define all(x) (x).begin(), (x).end() #define all1(x) (x).begin()+1, (x).end() #define loop(i, n) for(int i = 0; i < n; i++) #define loop1(i, n) for(int i = 1; i <= n; i++) #define print(x) cout << #x << ": " << x << endl << flush typedef long long ll; typedef vector<int> vi; typedef vector<double> vd; typedef array<int, 2> ii; typedef array<int, 3> ti; typedef vector<ii> vii; typedef vector<ti> vti; typedef vector<vi> vvi; typedef priority_queue<int> pq; template<class T> bool ckmin(T&a, T b) { bool B = a > b; a = min(a, b); return B; } template<class T> bool ckmax(T&a, T b) { bool B = a < b; a = max(a, b); return B; } const int inf = 1e17; double cst(int x) { return (double) x * (x-1) / 4; } void solve() { string s; cin >> s; int n = s.size(); vvi pos(26); loop(i, n) pos[s[i]-'A'].pb(i); vi ch; loop(i, 26) if(!pos[i].empty()) ch.pb(i); int g = ch.size(); vi id(26); loop(i, g) id[ch[i]] = i; if(g == 1) { cout << setprecision(16) << cst(n/2) + cst((n+1)/2) << endl; return; } vd dp(1 << g, inf); dp[0] = 0; auto cost = [&] (int pre, int ch, int idx) -> double { bitset<15> prev(pre); double ans = 0; int cnt = 0, oth = 0; for(int i = 0; i <= idx; i++) { if(prev[id[s[i]-'A']]) oth++; else if(id[s[i]-'A'] == ch) { cnt++; ans += oth; } } ans += cst(cnt); cnt = 0, oth = 0; for(int i = n-1; i > idx; i--) { if(prev[id[s[i]-'A']]) oth++; else if(id[s[i]-'A'] == ch) { cnt++; ans += oth; } } ans += cst(cnt); return ans; }; loop(pre, 1 << g) { bitset<15> prev(pre); loop(ch, g) if(!prev[ch]) { int nxt = pre + (1 << ch); int l = 0; int r = n-1; while(r - l >= 3) { int m1 = l + (r-l)/3; int m2 = l + (r-l)/3 * 2; double a = cost(pre, ch, m1); double b = cost(pre, ch, m2); if(a < b) r = m2; else l = m1; } for(int idx = l; idx <= r; idx++) ckmin(dp[nxt], dp[pre] + cost(pre, ch, idx)); } } cout << setprecision(16) << dp[(1 << g) - 1] << endl; } signed main() { fast_io; int t = 1; //cin >> t; while(t--) solve(); return 0; }
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