[P3453] [POI2007] DRZ - Trees
毫无技术含量的题目。
首先，特判 \(1\) 和 \(n\) 的答案，还有相邻的贡献。
然后记 \(l_i = min(a_{i - 1}, a_{i + 1})\)，记 \(r_i = max(a_{i - 1}, a_{i + 1})\)，记 \(s_i = |a_i - a_{i - 1}| + |a_i - a_{i + 1}|\)，考虑 \(i\) 和 \(j\) 所作贡献（\(i, j \in [2,n], |i - j| > 1\)）：
（1）\(a_i < l_j\)，\(a_j < l_i\)
\(\delta = l_i + r_i - 2a_i - s_i + l_j + r_j - 2a_j - s_j\)
（2）\(l_j \le a_i < r_j\)，\(a_j < l_i\)
\(\delta = l_i + r_i - s_i + r_j - l_j - 2a_j - s_j\)
（3）\(a_i \ge r_j\)，\(a_j < l_i\)
\(\delta = l_i + r_i + 2a_i - s_i - l_j - r_j - 2a_j - s_j\)
（4）\(a_i < l_j\)，\(l_i \le a_j < r_i\)
\(\delta = r_i - l_i - 2a_i - s_i + l_j + r_j - s_j\)
（5）\(l_j \le a_i < r_j\)，\(l_i \le a_j < r_i\)
\(\delta = r_i - l_i - s_i + r_j - l_j - s_j\)
（6）\(a_i \ge r_j\)，\(l_i \le a_j < r_i\)
\(\delta = r_i - l_i + 2a_i - s_i - l_j - r_j - s_j\)
（7）\(a_i < l_j\)，\(a_j \ge r_i\)
\(\delta = -l_i - r_i - 2a_i - s_i + l_j + r_j + 2a_j - s_j\)
（8）\(l_j \le a_i < r_j\)，\(a_j \ge r_i\)
\(\delta = -l_i - r_i - s_i + r_j - l_j + 2a_j - s_j\)
（9）\(a_i \ge r_j\)，\(a_j \ge r_i\)
\(\delta = -l_i - r_i + 2a_i - s_i - l_j - r_j + 2a_j - s_j\)
把 \(j\) 看作修改，把 \(i\) 看作询问。
\(j\) 对应的修改看成 \((l, a_j)\) 到 \((r, a_j)\) 的线段（限制为 \(l \le a_i \le r\)）。
\(i\) 对应的询问看成 \((a_i, l')\) 到 \((a_i, r')\) 的线段（限制为 \(l' \le a_j \le r'\)）。
两个线段有交即产生贡献，直接对 \(x\) 轴扫描线即可。
为了满足 \(|i - j| > 1\)，所以首先离散化的时候不应该去重，然后询问的时候特判 \(a_{i - 1}\) 和 \(a_{i + 1}\) 两个纵坐标不计算贡献。\(a_i\) 计算贡献无所谓，因为其贡献为 \(0\)。
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MAXN = 5e4 + 10, INF = numeric_limits<int>::max();
int n, a[MAXN], s[MAXN], id[MAXN], rev[MAXN], sz;
pair<int, int> bd[MAXN];
ll ans[MAXN], sans;
vector<pair<int, int>> op[MAXN];
inline void chsmin(ll &x, const ll &y) { y < x && (x = y); }
ll calc(int x, int y) {
ll ret = 0;
if (x != 1) ret += abs(a[y] - a[x - 1]) - abs(a[x] - a[x - 1]);
if (y != n) ret += abs(a[x] - a[y + 1]) - abs(a[y] - a[y + 1]);
if (x + 1 != y) {
ret += abs(a[x] - a[y - 1]) - abs(a[y] - a[y - 1]) +
abs(a[y] - a[x + 1]) - abs(a[x] - a[x + 1]);
}
return ret;
}
namespace sgt {
int mn[MAXN * 4];
inline int ls(int p) { return p << 1; }
inline int rs(int p) { return p << 1 | 1; }
inline void push_up(int p) { mn[p] = min(mn[ls(p)], mn[rs(p)]); }
void build(int p, int l, int r) {
mn[p] = INF;
if (l == r) return;
int mid = (l + r) >> 1;
build(ls(p), l, mid);
build(rs(p), mid + 1, r);
}
void update(int p, int l, int r, int x, int v) {
if (l == r) {
mn[p] = v;
return;
}
int mid = (l + r) >> 1;
if (x <= mid) {
update(ls(p), l, mid, x, v);
} else {
update(rs(p), mid + 1, r, x, v);
}
push_up(p);
}
int query(int p, int l, int r, int ql, int qr) {
if (ql > qr) return INF;
if (ql <= l && r <= qr) return mn[p];
int mid = (l + r) >> 1;
if (qr <= mid) {
return query(ls(p), l, mid, ql, qr);
} else if (ql > mid) {
return query(rs(p), mid + 1, r, ql, qr);
} else {
return min(query(ls(p), l, mid, ql, qr),
query(rs(p), mid + 1, r, ql, qr));
}
}
} // namespace sgt
int query(int p, int l, int r) {
int x = bd[p].first, y = bd[p].second;
if (r == y) --r;
if (r == x) --r;
if (l == x) ++l;
if (l == y) ++l;
if (l > r) return INF;
if ((x < l && r < y) || r < x || l > y) return sgt::query(1, 1, n, l, r);
if (l < x && r > y) {
return min({sgt::query(1, 1, n, l, x - 1),
sgt::query(1, 1, n, x + 1, y - 1),
sgt::query(1, 1, n, y + 1, r)});
}
if (l < x) {
return min(sgt::query(1, 1, n, l, x - 1),
sgt::query(1, 1, n, x + 1, r));
}
return min(sgt::query(1, 1, n, l, y - 1), sgt::query(1, 1, n, y + 1, r));
}
void solve1() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i) {
op[1].emplace_back(id[i], a[i - 1] + a[i + 1] - a[i] * 2 - s[i]);
op[bd[i].first].emplace_back(id[i], INF);
}
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p], (ll)query(p, 1, bd[p].first - 1) + a[p - 1] +
a[p + 1] - a[p] * 2 - s[p]);
}
}
}
void solve2() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i) {
op[bd[i].first].emplace_back(
id[i], abs(a[i - 1] - a[i + 1]) - a[i] * 2 - s[i]);
op[bd[i].second].emplace_back(id[i], INF);
}
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p], (ll)query(p, 1, bd[p].first - 1) + a[p - 1] +
a[p + 1] - s[p]);
}
}
}
void solve3() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i) {
op[bd[i].second].emplace_back(id[i],
-a[i - 1] - a[i + 1] - a[i] * 2 - s[i]);
}
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p], (ll)query(p, 1, bd[p].first - 1) + a[p - 1] +
a[p + 1] + a[p] * 2 - s[p]);
}
}
}
void solve4() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i) {
op[1].emplace_back(id[i], a[i - 1] + a[i + 1] - s[i]);
op[bd[i].first].emplace_back(id[i], INF);
}
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p], (ll)query(p, bd[p].first, bd[p].second - 1) +
abs(a[p - 1] - a[p + 1]) - a[p] * 2 - s[p]);
}
}
}
void solve5() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i) {
op[bd[i].first].emplace_back(id[i], abs(a[i - 1] - a[i + 1]) - s[i]);
op[bd[i].second].emplace_back(id[i], INF);
}
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p], (ll)query(p, bd[p].first, bd[p].second - 1) +
abs(a[p - 1] - a[p + 1]) - s[p]);
}
}
}
void solve6() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i)
op[bd[i].second].emplace_back(id[i], -a[i - 1] - a[i + 1] - s[i]);
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p], (ll)query(p, bd[p].first, bd[p].second - 1) +
abs(a[p - 1] - a[p + 1]) + a[p] * 2 - s[p]);
}
}
}
void solve7() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i) {
op[1].emplace_back(id[i], a[i - 1] + a[i + 1] + a[i] * 2 - s[i]);
op[bd[i].first].emplace_back(id[i], INF);
}
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p], (ll)query(p, bd[p].second, n) - a[p - 1] - a[p + 1] -
a[p] * 2 - s[p]);
}
}
}
void solve8() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i) {
op[bd[i].first].emplace_back(
id[i], abs(a[i - 1] - a[i + 1]) + a[i] * 2 - s[i]);
op[bd[i].second].emplace_back(id[i], INF);
}
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p],
(ll)query(p, bd[p].second, n) - a[p - 1] - a[p + 1] - s[p]);
}
}
}
void solve9() {
for (int i = 1; i <= n; ++i) op[i].clear();
for (int i = 2; i < n; ++i) {
op[bd[i].second].emplace_back(id[i],
-a[i - 1] - a[i + 1] + a[i] * 2 - s[i]);
}
sgt::build(1, 1, n);
for (int i = 1; i <= n; ++i) {
for (auto &j : op[i]) sgt::update(1, 1, n, j.first, j.second);
int p = rev[i];
if (p != 1 && p != n) {
chsmin(ans[p], (ll)query(p, bd[p].second, n) - a[p - 1] - a[p + 1] +
a[p] * 2 - s[p]);
}
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n;
for (int i = 1; i <= n; ++i) cin >> a[i];
iota(rev + 1, rev + n + 1, 1);
sort(rev + 1, rev + n + 1, [](int x, int y) { return a[x] < a[y]; });
for (int i = 1; i <= n; ++i) id[rev[i]] = i;
for (int i = 1; i < n; ++i) sans += abs(a[i] - a[i + 1]);
for (int i = 2; i <= n; ++i) chsmin(ans[1], calc(1, i));
for (int i = 1; i < n; ++i) chsmin(ans[n], calc(i, n));
for (int i = 2; i <= n; ++i) {
chsmin(ans[i],
min({calc(1, i), calc(i - 1, i), calc(i, i + 1), calc(i, n)}));
}
for (int i = 2; i < n; ++i) {
s[i] = abs(a[i + 1] - a[i]) + abs(a[i] - a[i - 1]);
bd[i] = minmax(id[i - 1], id[i + 1]);
}
solve1();
solve2();
solve3();
solve4();
solve5();
solve6();
solve7();
solve8();
solve9();
for (int i = 1; i <= n; ++i) cout << ans[i] + sans << "\n";
return 0;
}