分类讨论 + 数据结构好题。

以下设 $s_i =|a_i - a_{i-1}|+|a_i-a_{i+1}|$，$l_i = \min(a_{i-1},a_{i+1})$，$r_i = \max(a_{i-1},a_{i+1})$。

首先考虑每次交换两个数的可能方案。这里可以讨论绝对值拆开的情况。

比如说第一种情况，即 $a_i < l_j < r_j$，$a_j<r_i<l_i$ 时（令 $i,j$ 为交换的元）。此时的答案偏移量明显为 $l_j+r_j-2a_i+l_i+r_i-2a_j-s_i-s_j$。其他情况可同理推出。

然后此时就可以用扫描线，即将 $j$ 当做线段修改，$i$ 当做询问。

需要注意的是，$1$ 和 $n$ 的情况需要单独考虑，相邻值交换的影响也可以单独直接计算出。

#include <cmath>
#include <vector>
#include <cstdio>
#include <utility>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;
typedef pair<int, int> pii;
#define lc (x << 1)
#define rc (x << 1 | 1)
#define mpr make_pair
#define pb push_back
#define px first
#define py second

const int N = 50005;
const int inf = 1e18;
int n, ori; 
int a[N], id[N], rev[N], ans[N], s[N], bl[N], br[N];
vector<pii> vc[N];

namespace sgtree{
	struct segtree{
		int l, r, mn;
		#define l(x) tree[x].l
		#define r(x) tree[x].r
		#define mn(x) tree[x].mn
	};
	segtree tree[N << 2];
	void pushup(int x){mn(x) = min(mn(lc), mn(rc));}
	void build(int x, int l, int r){
		mn(x) = inf; l(x) = l; r(x) = r;
		if(l == r) return; int mid = (l + r) >> 1;
		build(lc, l, mid); build(rc, mid + 1, r);
		pushup(x);
	}
	void modify(int x, int p, int val){
		if(l(x) == r(x)){mn(x)= val; return;}
		int mid = (l(x) + r(x)) >> 1;
		if(p <= mid) modify(lc, p, val); else modify(rc, p, val);
		pushup(x);
	}
	int query(int x, int l, int r){
		if(l <= l(x) && r(x) <= r) return mn(x);
		int mid = (l(x) + r(x)) >> 1;
		if(r <= mid) return query(lc, l, r);
		if(l > mid) return query(rc, l, r);
		return min(query(lc, l, r), query(rc, l, r)); 
	}
};
using namespace sgtree;

int ask(int p, int l, int r){
	int x = bl[p], y = br[p];
	if(r == y) -- r; if(r == x) -- r;
	if(l == x) ++ l; if(l == y) ++ l;
	if(l > r) return inf;
	return query(1, l, r);
	if((x < l && r < y) || r < x || l > y) return query(1, l, r);
	if(l < x && r > y) return min(query(1, l, x - 1), min(query(1, x + 1, y - 1), query(1, y + 1, r)));
	if(l < x) return min(query(1, l, x - 1), query(1, x + 1, r));
	return min(query(1, l, y - 1), query(1, y + 1, r));
}
void solve1(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++){
		vc[1].pb(mpr(id[i], a[i - 1] + a[i + 1] - 2 * a[i] - s[i]));
		vc[bl[i]].pb(mpr(id[i], inf));
	}
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n)
		  ans[p] = min(ans[p], ask(p, 1, bl[p] - 1) + a[p - 1] + a[p + 1] - 2 * a[p] - s[p]);
	}
}
void solve2(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++){
		vc[bl[i]].pb(mpr(id[i], abs(a[i - 1] - a[i + 1]) - 2 * a[i] - s[i]));
		vc[br[i]].pb(mpr(id[i], inf));
	}
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n){
			ans[p] = min(ans[p], ask(p, 1, bl[p] - 1) + a[p - 1] + a[p + 1] - s[p]);
		}
	}
}
void solve3(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++)
		vc[br[i]].pb(mpr(id[i], - 2 * a[i] - a[i - 1] - a[i + 1] - s[i]));
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n)
		  ans[p] = min(ans[p], ask(p, 1, bl[p] - 1) + a[p - 1] + a[p + 1] + 2 * a[p] - s[p]);
	}
}
void solve4(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++){
		vc[1].pb(mpr(id[i], a[i - 1] + a[i + 1] - s[i]));
		vc[bl[i]].pb(mpr(id[i], inf));
	}
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n)
		  ans[p] = min(ans[p], ask(p, bl[p], br[p] - 1) + abs(a[p + 1] - a[p - 1]) - 2 * a[p] - s[p]);
	}
}
void solve5(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++){
		vc[bl[i]].pb(mpr(id[i], abs(a[i + 1] - a[i - 1]) - s[i]));
		vc[br[i]].pb(mpr(id[i], inf));
	}
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n)
		  ans[p] = min(ans[p], ask(p, bl[p], br[p] - 1) + abs(a[p + 1] - a[p - 1]) - s[p]);
	}
}
void solve6(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++)
		vc[br[i]].pb(mpr(id[i], - a[i + 1] - a[i - 1] - s[i]));
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n)
		  ans[p] = min(ans[p], ask(p, bl[p], br[p] - 1) + abs(a[p + 1] - a[p - 1]) + 2 * a[p] - s[p]);
	}
}
void solve7(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++){
		vc[1].pb(mpr(id[i], a[i + 1] + a[i - 1] + 2 * a[i] - s[i]));
		vc[bl[i]].pb(mpr(id[i], inf));
	}
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n)
		  ans[p] = min(ans[p], ask(p, br[p], n) - a[p - 1] - a[p + 1] - 2 * a[p] - s[p]);
	}
}
void solve8(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++){
		vc[bl[i]].pb(mpr(id[i], abs(a[i + 1] - a[i - 1]) + 2 * a[i] - s[i]));
		vc[br[i]].pb(mpr(id[i], inf));
	}
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n)
		  ans[p] = min(ans[p], ask(p, br[p], n) - a[p - 1] - a[p + 1] - s[p]);
	}
}
void solve9(){
	for(int i = 1; i <= n; i ++) vc[i].clear(); build(1, 1, n);
	for(int i = 2; i < n; i ++)
		vc[br[i]].pb(mpr(id[i], 2 * a[i] - a[i + 1] - a[i - 1] - s[i]));
	for(int i = 1; i <= n; i ++){
		for(auto k: vc[i]) modify(1, k.px, k.py);
		int p = rev[i];
		if(p != 1 && p != n)
		  ans[p] = min(ans[p], ask(p, br[p], n) + 2 * a[p] - a[p - 1] - a[p + 1] - s[p]);
	}
}
int calc(int x, int y){
	if(x > y) swap(x, y);
	int res = 0;
	if(x != 1) res += abs(a[y] - a[x - 1]) - abs(a[x] - a[x - 1]);
	if(y != n) res += abs(a[x] - a[y + 1]) - abs(a[y] - a[y + 1]);
	if(x + 1 != y) res += abs(a[y] - a[x + 1]) + abs(a[x] - a[y - 1]) - abs(a[x + 1] - a[x]) - abs(a[y] - a[y - 1]);
	return res;
}

signed main(){
	scanf("%lld", &n);
	for(int i = 1; i <= n; i ++) scanf("%lld", &a[i]), rev[i] = i;
	for(int i = 2; i <= n; i ++) ori += abs(a[i] - a[i - 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 = 2; i <= n; i ++) ans[1] = min(ans[1], calc(1, i));
	for(int i = 1; i < n; i ++) ans[n] = min(ans[n], calc(i, n)); 
	for(int i = 2; i < n; i ++){
		ans[i] = min(ans[i], min(min(calc(1, i), calc(i, n)), min(calc(i, i - 1), calc(i, i + 1))));
		s[i] = abs(a[i] - a[i - 1]) + abs(a[i + 1] - a[i]);
		bl[i] = min(id[i - 1], id[i + 1]); br[i] = max(id[i - 1], id[i + 1]);
	}
	solve1(); solve2(); solve3();
	solve4(); solve5(); solve6();
	solve7(); solve8(); solve9();
	for(int i = 1; i <= n; i ++) printf("%lld\n", ori + ans[i]); return 0; 
}