Submission #284361

#TimeUsernameProblemLanguageResultExecution timeMemory
284361_7_7_Boxes with souvenirs (IOI15_boxes)C++14
20 / 100
1 ms384 KiB
#include "boxes.h" #include <bits/stdc++.h> #include <ext/pb_ds/assoc_container.hpp> #include <ext/pb_ds/tree_policy.hpp> using namespace std; using namespace __gnu_pbds; //#define int long long //#pragma GCC optimize("Ofast") //#pragma comment(linker, "/stack:200000000") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4") #define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout); #define fastio ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); #define all(x) x.begin(), x.end() #define sz(s) (int)s.size() #define pb push_back #define ppb pop_back #define mp make_pair #define s second #define f first typedef pair < long long, long long > pll; typedef pair < int, int > pii; typedef unsigned long long ull; typedef vector < pii > vpii; typedef vector < int > vi; typedef long double ldb; typedef long long ll; typedef double db; typedef tree < int, null_type, less < int >, rb_tree_tag, tree_order_statistics_node_update > ordered_set; const int inf = 1e9, maxn = 2e5 + 48, mod = 998244353, N = 1e7 + 12; const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 300; const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9); const db eps = 1e-12, pi = acos(-1); const ll INF = 1e18; ll s1[N], s2[N], p1[N], p2[N]; int pos[N]; ll delivery(int n, int k, int l, int p[]) { sort(p, p + n); int cur = 0, cnt = 0, j = 0; while (j < n && !p[j]) ++j; if (j == n) return 0; for (int i = j; i < n; ++i) { p1[i] = p[i]; p2[i] = l - p[i]; } for (int i = j; i < n; ++i) { if (cnt == k) { cnt = 0; pos[cur++] = i - 1; } ++cnt; if (!cur) s1[i] = p1[i] + min(p1[i], p2[i]); else s1[i] = s1[pos[cur - 1]] + p1[i] + min(p1[i], p2[i]); } cur = 0, cnt = 0; for (int i = n - 1; i >= j; --i) { if (cnt == k) { cnt = 0; pos[cur++] = i + 1; } ++cnt; if (!cur) s2[i] = p2[i] + min(p1[i], p2[i]); else s2[i] = s2[pos[cur - 1]] + p2[i] + min(p1[i], p2[i]); } // for (int i = 0; i < n; ++i) // cerr << s1[i] << ' ' << s2[i] << endl; ll ans = min(s1[n - 1], s2[j]); for (int i = j; i < n - 1; ++i) ans = min(ans, s1[i] + s2[i + 1]); return ans; }
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...