제출 #120785

#제출 시각아이디문제언어결과실행 시간메모리
120785youngyojun휴가 (IOI14_holiday)C++11
100 / 100
306 ms45448 KiB
#include "holiday.h" #include <bits/stdc++.h> #define upmax(a,b) (a)=max((a),(b)) #define INFLL (0x3f3f3f3f3f3f3f3fll) using namespace std; typedef long long ll; const int MAXN = 100055; struct NOD { NOD() : sum(0), cnt(0), l(0), r(0) {} ll sum; int cnt, l, r; } arr[MAXN*18]; int arrn; int newNOD() { return ++arrn; } int pst[MAXN]; int A[MAXN], AO[MAXN], ARO[MAXN]; ll Ans; int N, X, L; void makeTree(int bf, int nw, int s, int e, int x, int r) { arr[nw].cnt = arr[bf].cnt + 1; arr[nw].sum = arr[bf].sum + r; if(s == e) return; int m = (s+e) >> 1; if(x <= m) { arr[nw].r = arr[bf].r; arr[nw].l = newNOD(); makeTree(arr[bf].l, arr[nw].l, s, m, x, r); } else { arr[nw].l = arr[bf].l; arr[nw].r = newNOD(); makeTree(arr[bf].r, arr[nw].r, m+1, e, x, r); } } ll _get(int bf, int nw, int x) { if(!x) return 0; if(arr[nw].cnt - arr[bf].cnt == x) return arr[nw].sum - arr[bf].sum; int lc = arr[arr[nw].l].cnt - arr[arr[bf].l].cnt; if(x <= lc) return _get(arr[bf].l, arr[nw].l, x); return (arr[arr[nw].l].sum - arr[arr[bf].l].sum) + _get(arr[bf].r, arr[nw].r, x-lc); } ll get(int s, int e, int x) { return _get(pst[s-1], pst[e], x); } void f(int s, int e, int p, int q) { if(s > e || p > q) return; int m = (s+e) >> 1; ll hc = -INFLL; int hi = -1; for(int x = p, xe = min({N, (L+X+m)/2, q}); x <= xe; x++) { ll t = get(m, x, min(x-m+1, L-(x+x-m-X))); if(t <= hc) continue; hc = t; hi = x; } upmax(Ans, hc); f(s, m-1, p, hi); f(m+1, e, hi, q); } ll getAns() { iota(AO, AO+N+1, 0); sort(AO+1, AO+N+1, [&](int a, int b) { return A[a] > A[b]; }); for(int i = 1; i <= N; i++) ARO[AO[i]] = i; for(int i = 1; i <= N; i++) { pst[i] = newNOD(); makeTree(pst[i-1], pst[i], 1, N, ARO[i], A[i]); } f(max(1, X-L), X, X, N); arrn = 0; X = N+1-X; reverse(A+1, A+N+1); iota(AO, AO+N+1, 0); sort(AO+1, AO+N+1, [&](int a, int b) { return A[a] > A[b]; }); for(int i = 1; i <= N; i++) ARO[AO[i]] = i; for(int i = 1; i <= N; i++) { pst[i] = newNOD(); makeTree(pst[i-1], pst[i], 1, N, ARO[i], A[i]); } f(max(1, X-L), X, X, N); return Ans; } long long int findMaxAttraction(int n, int start, int d, int attraction[]) { ::N = n; ::X = start + 1; ::L = d; for(int i = 0; i < n; i++) ::A[i+1] = attraction[i]; return getAns(); }
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