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题解:P13523 [KOI 2025 #2] 序列与查询
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Description
给定序列 和 次查询。
每次查询给出一个数 ,求:
Limitations
Solution
但这里 可以为负。由于本题可以离线,我们将所有询问按照 升序排序,第一次询问单独做,然后遍历剩下的询问,每次全局加 然后查全局
GSS 就是答案,显然每次加的都是正数,那么就可以了。时间复杂度 ,稍微优化一下常数即可通过。
code
CPP#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
using ui64 = unsigned long long;
using i128 = __int128;
using ui128 = unsigned __int128;
using f4 = float;
using f8 = double;
using f16 = long double;
template<class T>
bool chmax(T &a, const T &b){
if(a < b){ a = b; return true; }
return false;
}
template<class T>
bool chmin(T &a, const T &b){
if(a > b){ a = b; return true; }
return false;
}
namespace Fastio {}
using Fastio::qin;
using Fastio::qout;
constexpr i64 inf = 4e18;
struct line {
int k; i64 b;
inline line() : line(0, 0) {}
inline line(int _k, i64 _b): k(_k), b(_b) {}
inline void add(i64 v) { b += k * v; }
};
inline line operator+(const line& lhs, const line& rhs) {
return line(lhs.k + rhs.k, lhs.b + rhs.b);
}
inline pair<line, i64> _max(const line& a, const line& b) {
if (a.k < b.k || (a.k == b.k && a.b < b.b)) return _max(b, a);
if (a.b >= b.b) return make_pair(a, inf);
return make_pair(b, (b.b - a.b) / (a.k - b.k));
}
struct info {
line pre, suf, ans, sum;
i64 x;
inline info() {}
inline info(line pre, line suf, line ans, line sum, i64 x)
: pre(pre), suf(suf), ans(ans), sum(sum), x(x) {}
};
inline info operator+(const info& a, const info& b) {
i64 x0 = min(a.x, b.x);
line sum = a.sum + b.sum;
auto [pre, x1] = _max(a.pre, a.sum + b.pre);
auto [suf, x2] = _max(b.suf, a.suf + b.sum);
auto [tmp, x3] = _max(a.ans, b.ans);
auto [ans, x4] = _max(tmp, a.suf + b.pre);
return info(pre, suf, ans, sum, min({x0, x1, x2, x3, x4}));
}
inline void operator+=(info& a, i64 v) {
a.x -= v;
a.pre.add(v), a.suf.add(v);
a.sum.add(v), a.ans.add(v);
}
struct node {
int l, r;
info dat;
i64 tag;
};
inline int ls(int u) { return 2 * u + 1; }
inline int rs(int u) { return 2 * u + 2; }
struct ktt {
vector<node> tr;
inline ktt() {}
inline ktt(const vector<i64>& a) {
const int n = a.size();
tr.resize(n << 2);
build(0, 0, n - 1, a);
}
inline void pushup(int u) { tr[u].dat = tr[ls(u)].dat + tr[rs(u)].dat; }
inline void apply(int u, i64 v) { tr[u].tag += v, tr[u].dat += v; }
inline void pushdown(int u) {
if (tr[u].tag) {
apply(ls(u), tr[u].tag);
apply(rs(u), tr[u].tag);
tr[u].tag = 0;
}
}
inline void build(int u, int l, int r, const vector<i64>& a) {
tr[u].l = l, tr[u].r = r;
if (l == r) {
line f(1, a[l]);
tr[u].dat = info(f, f, f, f, inf);
return;
}
const int mid = (l + r) >> 1;
build(ls(u), l, mid, a);
build(rs(u), mid + 1, r, a);
pushup(u);
}
inline void defeat(int u, i64 v) {
const int mid = (tr[u].l + tr[u].r) >> 1;
if (v > tr[u].dat.x) {
defeat(ls(u), tr[u].tag + v);
defeat(rs(u), tr[u].tag + v);
tr[u].tag = 0;
pushup(u);
}
else apply(u, v);
}
};
struct query {
i64 x; int id;
inline query() {}
inline query(i64 x, int id) : x(x), id(id) {}
inline bool operator<(const query& b) const {
return x < b.x;
}
};
signed main() {
ios::sync_with_stdio(0);
cin.tie(0), cout.tie(0);
int n, q;
qin >> n >> q;
vector<i64> a(n);
for (int i = 0; i < n; i++) qin >> a[i];
vector<query> que(q);
for (int i = 0; i < q; i++) qin >> que[i].x, que[i].id = i;
sort(que.begin(), que.end());
for (int i = 0; i < n; i++) a[i] += que[0].x;
ktt tree(a);
vector<i64> ans(q);
ans[que[0].id] = tree.tr[0].dat.ans.b;
for (int i = 1; i < q; i++) {
tree.defeat(0, que[i].x - que[i - 1].x);
ans[que[i].id] = tree.tr[0].dat.ans.b;
}
for (int i = 0; i < q; i++) cout << ans[i] << '\n';
return 0;
}
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