제출 #1287671

#제출 시각아이디문제언어결과실행 시간메모리
1287671KickingKun3단 점프 (JOI19_jumps)C++20
100 / 100
476 ms91988 KiB
#include <bits/stdc++.h> using namespace std; #define ll long long #define ull unsigned long long #define ld long double #define bigint __int128 #define emb emplace_back #define pb push_back #define pii pair <int, int> #define fi first #define se second #define all(v) v.begin(), v.end() #define Task "" #define MASK(k) (1ull << k) #define bitcnt(k) __builtin_popcount(k) #define testBit(n, k) ((n >> k) & 1) #define flipBit(n, k) (n ^ (1ll << k)) #define offBit(n, k) (n & ~MASK(k)) #define onBit(n, k) (n | (1ll << k)) template <class T> bool minimize(T &a, T b) {if (a > b) {a = b; return true;} return false;} template <class T> bool maximize(T &a, T b) {if (a < b) {a = b; return true;} return false;} const int N = 5e5 + 5, lim = 60, mod = 1e9 + 7; const ll INF = 1e18; int n, q, a[N]; int l[N], r[N]; namespace sub2 { int f[5005][5005]; void solve() { for (int x = 1; x <= n - 2; x++) { for (int z = x + 2, y = x + 1, ma = a[y]; z <= n; z++) { while ((y + 1) * 2 <= x + z) maximize(ma, a[++y]); f[x][z] = a[x] + ma + a[z]; } } for (int len = 4; len <= n; len++) for (int x = 1, z = len; z <= n; x++, z++) maximize(f[x][z], max(f[x][z - 1], f[x + 1][z])); for (int i = 1; i <= q; i++) cout << f[l[i]][r[i]] << '\n'; } } namespace sub3 { int sufMa[N]; vector <pii> candidate; void solve() { for (int i = n; i > 0; i--) sufMa[i] = max(sufMa[i + 1], a[i]); // nhung cap (a, b) : max(A[a+1, b-1]) < min(A[a], A[b]) stack <int> st; for (int i = 1; i <= n; i++) { while (!st.empty() && a[st.top()] <= a[i]) { // neu truoc i ton tai k ma A[x] <= A[k] -> x da bi bo / neu ko thi A[x] lon nhat // den khi A[x] <= A[i] -> x va i la 2 dau mut lon nhat candidate.emplace_back(st.top(), i); st.pop(); } if (!st.empty()) { // A[st.top()] > A[st.top() + 1 ... i] // cap nay co the chac chan thoa khi A[st.top] > A[i] > max o giua // nhung vi tri trong st con lai ko thoa vi co max_mid >= A[st.top()] > A[i] candidate.emplace_back(st.top(), i); } st.emplace(i); } int ans = 0; for (auto [x, y]: candidate) { int miZ = 2 * y - x; if (miZ <= n) maximize(ans, a[x] + a[y] + sufMa[miZ]); } cout << ans; } } namespace sub4 { struct Node { int mxF, mxA, mxAns; Node () {} Node (int a, int b, int c) { mxF = a; mxA = b; mxAns = c; } // l <= i <= j <= r -> mxAns = max(Fi + Aj) Node operator + (const Node &other) { Node res; res.mxF = max(mxF, other.mxF); res.mxA = max(mxA, other.mxA); res.mxAns = max({mxAns, other.mxAns, mxF + other.mxA}); return res; } }; struct SegmentTree { vector <Node> node; SegmentTree (int len) { node.assign(len * 4 + 1, Node()); } void build(int id, int l, int r) { if (l == r) { node[id] = Node(0, a[l], a[l]); return; } int mid = (l + r) >> 1; build(id << 1, l, mid); build(id << 1 | 1, mid + 1, r); node[id] = node[id << 1] + node[id << 1 | 1]; } void update(int id, int l, int r, int p, int f) { if (l == r) { maximize(f, node[id].mxF); node[id] = Node(f, a[l], f + a[l]); return; } int mid = (l + r) >> 1; if (p <= mid) update(id << 1, l, mid, p, f); else update(id << 1 | 1, mid + 1, r, p, f); node[id] = node[id << 1] + node[id << 1 | 1]; } Node get(int id, int l, int r, int u, int v) { if (u > r || v < l) return Node(0, 0, 0); if (u <= l && r <= v) return node[id]; int mid = (l + r) >> 1; return get(id << 1, l, mid, u, v) + get(id << 1 | 1, mid + 1, r, u, v); } }; vector <int> candidate[N]; vector <pii> query[N]; void solve() { stack <int> st; for (int i = 1; i <= n; i++) { while (!st.empty() && a[st.top()] <= a[i]) { // neu truoc i ton tai k ma A[x] <= A[k] -> x da bi bo / neu ko thi A[x] lon nhat // den khi A[x] <= A[i] -> x va i la 2 dau mut lon nhat candidate[st.top()].emplace_back(i); st.pop(); } if (!st.empty()) { // A[st.top()] > A[st.top() + 1 ... i] // cap nay co the chac chan thoa khi A[st.top] > A[i] > max o giua // nhung vi tri trong st con lai ko thoa vi co max_mid >= A[st.top()] > A[i] candidate[st.top()].emplace_back(i); } st.emplace(i); } for (int i = 1; i <= q; i++) { query[l[i]].emplace_back(r[i], i); } SegmentTree ST(n); vector <int> ans(q + 1); ST.build(1, 1, n); for (int l = n; l > 0; l--) { for (int b: candidate[l]) if (2 * b - l <= n) { ST.update(1, 1, n, 2 * b - l, a[l] + a[b]); } for (auto [r, id]: query[l]) { ans[id] = ST.get(1, 1, n, l, r).mxAns; } } for (int i = 1; i <= q; i++) cout << ans[i] << '\n'; } } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); if (fopen(Task".inp", "r")) { freopen(Task".inp", "r", stdin); freopen(Task".out", "w", stdout); } cin >> n; for (int i = 1; i <= n; i++) cin >> a[i]; cin >> q; for (int i = 1; i <= q; i++) cin >> l[i] >> r[i]; if (n <= 5e3) sub2::solve(); else if (q == 1 && l[1] == 1 && r[1] == n) sub3::solve(); else sub4::solve(); }

컴파일 시 표준 에러 (stderr) 메시지

jumps.cpp: In function 'int main()':
jumps.cpp:188:24: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  188 |                 freopen(Task".inp", "r", stdin);
      |                 ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~
jumps.cpp:189:24: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  189 |                 freopen(Task".out", "w", stdout);
      |                 ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~
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