Submission #773190

#TimeUsernameProblemLanguageResultExecution timeMemory
773190GrindMachineCandies (JOI18_candies)C++17
100 / 100
481 ms30348 KiB
// Om Namah Shivaya #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; template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; typedef long long int ll; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; #define fastio ios_base::sync_with_stdio(false); cin.tie(NULL) #define pb push_back #define endl '\n' #define sz(a) a.size() #define setbits(x) __builtin_popcountll(x) #define ff first #define ss second #define conts continue #define ceil2(x, y) ((x + y - 1) / (y)) #define all(a) a.begin(), a.end() #define rall(a) a.rbegin(), a.rend() #define yes cout << "Yes" << endl #define no cout << "No" << endl #define rep(i, n) for(int i = 0; i < n; ++i) #define rep1(i, n) for(int i = 1; i <= n; ++i) #define rev(i, s, e) for(int i = s; i >= e; --i) #define trav(i, a) for(auto &i : a) template<typename T> void amin(T &a, T b) { a = min(a, b); } template<typename T> void amax(T &a, T b) { a = max(a, b); } #ifdef LOCAL #include "debug.h" #else #define debug(x) 42 #endif /* refs: edi of apio 2007 backup */ const int MOD = 1e9 + 7; const int N = 1e5 + 5; const int inf1 = int(1e9) + 5; const ll inf2 = ll(1e18) + 5; void solve(int test_case) { ll n; cin >> n; vector<ll> a(n+5); rep1(i,n) cin >> a[i]; set<array<ll,3>> segs; set<array<ll,3>> costs; rep1(i,n){ segs.insert({i,i+1,a[i]}); costs.insert({a[i],i,i+1}); } ll ans = 0; rep1(iter,ceil2(n,2)){ auto [mx,l,r] = *costs.rbegin(); ans += mx; cout << ans << endl; auto it = segs.lower_bound({l,r,mx}); array<ll,3> ar1, ar2; ar1.fill(-1); ar2.fill(-1); if(it != segs.begin()){ ar1 = *prev(it); } if(next(it) != segs.end()){ ar2 = *next(it); } if(ar1[0] != -1){ segs.erase(ar1); costs.erase({ar1[2], ar1[0], ar1[1]}); } if(ar2[0] != -1){ segs.erase(ar2); costs.erase({ar2[2], ar2[0], ar2[1]}); } if(ar1[0] != -1 and ar2[0] != -1){ ll l = ar1[0], r = ar2[1], cost = ar1[2]+ar2[2]-mx; segs.insert({l,r,cost}); costs.insert({cost,l,r}); } segs.erase({l,r,mx}); costs.erase({mx,l,r}); } } int main() { fastio; int t = 1; // cin >> t; rep1(i, t) { solve(i); } return 0; }
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