思路和各题解类似，但必须加上第17行才对。

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
const int N = 200010;

int T, n;
int l[N], r[N];
LL p[N], s[N], res[N]; 
struct node
{
	LL t, a, b;
	bool operator <(const node &x) const
	{
		if (t != x.t) return t < x.t;
		return a < x.a;
	}
};
set <node> S; 

LL calc(int a, int b)
{
	return (p[b] - p[a] + s[a] + s[b] - 1) / (s[a] + s[b]) * 2 - (a & 1); 
}

int main()
{
	scanf("%d", &T);
	while (T -- )
	{
		scanf("%d", &n);
		for (int i = 1; i <= n; i ++ ) scanf("%lld", &p[i]);
		for (int i = 1; i <= n; i ++ ) scanf("%lld", &s[i]);
		for (int i = 1; i <= n; i ++ ) l[i] = i - 1, r[i] = i + 1;
		for (int i = 1; i < n; i ++ ) S.insert({calc(i, i + 1), i, i + 1});
		while (S.size())
		{
			node tmp = *S.begin(); S.erase(S.begin());
			int a = tmp.a, b = tmp.b;
			res[a] = res[b] = tmp.t;
			if (l[a]) S.erase({calc(l[a], a), l[a], a});
			if (r[b] != n + 1) S.erase({calc(b, r[b]), b, r[b]});
			if (l[a] && r[b] != n + 1) S.insert({calc(l[a], r[b]), l[a], r[b]});
			r[l[a]] = r[b]; l[r[b]] = l[a];
		}
		for (int i = 1; i <= n; i ++ ) printf("%lld ", res[i]);
		printf("\n");
	}
	
	return 0;
}

若不加17行，我发现每删去相邻点$a$, $b$时, 消除l[a] 和r[b]所需时间有可能小于小于消除$a$, $b$的时间。这是为什么呢？有大佬能证明吗？