https://www.luogu.com.cn/problem/CF765F

挺不错的一道题，首先考虑扫描线，对于每个\(i\),找\(j<i,a[j]>a[i], j\)最大的

用权值线段树可以轻易维护

假设找到一个\(j\)，然后再找\((a[i],a[j])\)之间的

时间复杂度不优

考虑再找到的一个是\(j'\)

发现必须要满足\(a[j']-a[i]<a[j]-a[i]\) 且 \(a[j']-a[i]<a[j]-a[j']\)

综合一下如果\(j'\)能贡献答案的话就必须满足第二条式子

移动一下项可以得到

\(2a[j']<a[j]+a[i] \\ a[j']<\frac{a[j]+a[i]}{2}\)

直接找即可

然后再用一个树状数组维护后缀最大值

code:

#include<bits/stdc++.h>
#define N 300050
#define pii pair<int, int>
#define ls ch[rt][0]
#define rs ch[rt][1]
#define fi first
#define se second
using namespace std;
const int lim = 1e9;
int ma[N << 5], ch[N << 5][2], tot, root;
void add(int &rt, int l, int r, int x, int o) {
    if(!rt) rt = ++ tot;
    ma[rt] = max(ma[rt], o);
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(x <= mid) add(ls, l, mid, x, o);
    else add(rs, mid + 1, r, x, o); 
}
int query(int rt, int l, int r, int L, int R) {
    if(L > R) return 0;
    if(!rt) return 0;
    if(L <= l && r <= R) return ma[rt];
    int mid = (l + r) >> 1, ret = 0;
    if(L <= mid) ret = query(ls, l, mid, L, R);
    if(R > mid) ret = max(ret, query(rs, mid + 1, r, L, R));
    return ret;
}
#define lowbit(x) (x & - x)
int t[N], n, a[N];
void update(int x, int y) {
    for(; x; x -= lowbit(x)) t[x] = min(t[x], y);
}
int ask(int x) {
    int ret = lim;
    for(; x <= n; x += lowbit(x)) ret = min(ret, t[x]);
    return ret;
}
void clr() {
    for(int i = 0; i <= n; i ++) t[i] = lim;
    for(int i = 0; i <= tot; i ++) ch[i][0] = ch[i][1] = ma[i] = 0;
    tot = root = 0;
}
vector<pii > q[N];
int ans[N], m;
void solve() {
    clr();    
    for(int i = 1; i <= n; i ++) {
        int pos = query(root, 0, lim, a[i], lim);
        while(pos) {
            update(pos, a[pos] - a[i]);
            pos = query(root, 0, lim, a[i], (a[i] + a[pos] + 1) / 2 - 1);
        }
         add(root, 0, lim, a[i], i);
        for(auto it : q[i]) {
            ans[it.se] = min(ans[it.se], ask(it.fi));
        }
    }
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++) scanf("%d", &a[i]);
    scanf("%d", &m);
    for(int i = 1; i <= m; i ++) {
        int l, r;
        scanf("%d%d", &l, &r);
        q[r].push_back(make_pair(l, i));
    }
    for(int i = 1; i <= m; i ++) ans[i] = lim;

    solve();
    for(int i = 1; i <= n; i ++) a[i] = lim - a[i];
    solve();
    for(int i = 1; i <= m; i ++) printf("%d\n", ans[i]);
    return 0;
}