제출 #1031319

#제출 시각아이디문제언어결과실행 시간메모리
1031319amine_aroua서열 (APIO23_sequence)C++17
43 / 100
206 ms43864 KiB
#include "sequence.h" #include <bits/stdc++.h> using namespace std; struct Node { int sum , pfmax , pfmin; }; Node comb(Node a , Node b) { Node ans; ans.sum = a.sum + b.sum; ans.pfmax = max(a.sum + b.pfmax , a.pfmax); ans.pfmin = min(a.sum + b.pfmin , a.pfmin); return ans; } struct segTree { int BASE; vector<Node> tree; void init(int n) { BASE = n; while(__builtin_popcount(BASE) != 1) BASE++; tree.assign(2*BASE , {0 , 0 , 0}); for(int i = 0 ; i < n ; i++) { tree[i + BASE] = {-1 , 0 , -1}; } for(int i = BASE - 1 ; i >= 1 ; i--) { tree[i] = comb(tree[i<<1] , tree[i<<1|1]); } } void upd(int i , int x) { tree[i + BASE] = {x , max(0 , x) , min(0 , x)}; for(int j = (i + BASE) / 2 ; j >= 1 ; j>>=1) { tree[j] = comb(tree[j<<1] , tree[j<<1|1]); } } Node query(int l , int r) { if(r == -1) { return {0 , 0 , 0}; } l+=BASE; r+=BASE; if(l == r) return tree[l]; Node lt = tree[l], rt = tree[r]; while(l + 1 < r) { if(l%2 == 0) { lt = comb(lt , tree[l + 1]); } if(r%2 == 1) { rt = comb(tree[r - 1] , rt); } l>>=1; r>>=1; } return comb(lt , rt); } }tree; bool check(int i , int j , int w , int N) { auto rt = tree.query(j , N - 1); int tot = tree.query(0 , j - 1).sum; int mxj = rt.pfmax + tot; int mnj = rt.pfmin + tot; auto lt = tree.query(0 , i - 1); int mxi = lt.pfmax; int mni = lt.pfmin; int mx = mxj - mni; int mn = mnj - mxi; if(mn <= 0 && mx >= 0) { return 1; } else if(mn > 0) { if(w >= mn) return 1; return 0; } else { if(w >= abs(mx)) { return 1; } return 0; } return 0; } int sequence(int N, std::vector<int> A) { vector<int> occ[N + 1]; tree.init(N); for(int i = 0 ; i < N ; i++) { A[i]--; occ[A[i]].push_back(i); } int ans = 1; for(int cur = N - 1 ; cur >= 0 ; cur--) { for(int i = 0 ; i < (int)occ[cur + 1].size() ; i++) { tree.upd(occ[cur + 1][i], 1); } for(int i = 0 ; i < (int)occ[cur].size() ; i++) { tree.upd(occ[cur][i] , 0); } for(int i = (int)occ[cur].size() - 1 ; i >= 0 ; i--) { while(ans + i < (int)occ[cur].size() && check(occ[cur][i] , occ[cur][ans + i] , ans + 1 , N)) { ans++; } } } return ans; }
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