Title:

l+1) * a − I = l Σ r ci − gap (L, R), where gap(l,r)=max ⁡ i=l+1r(di − di − 1) 2gap (L, R) = \ Max \ limits {I = L + 1} ^ R (D_ i-d_{i-1})^2gap(l,r)=i=l+1maxr​(di​−di−1​)2. (n≤3×105)(n \leq 3×10^5)(n≤3×105)

Links:

https://codeforces.com/contest/1107/problem/G

Solutions:

In this paper, the author analyzes the characteristics of ], query the largest sum[r]sum[r]sum[r] on the right, and then subtract the smallest sum[l]sum[l]sum[l] on the left to get the maximum interval (or processing prefix and suffix sum) across the gap gap.

Reference code:

Sorting + parallel query + line tree:

#include<bits/stdc++.h>
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define sz(a) ((int)a.size())
#define pb push_back
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define gmid (l + r >> 1)
const int maxn = 3e5 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
 
typedef pair<ll, int> pli;
const ll oo = 1ll << 60;
pli tp[maxn];
ll d[maxn], c[maxn], sum[maxn];
int pre[maxn], li[maxn], ri[maxn];
int n, A; ll ans;
 
struct SegTree{
 
    ll mx[maxn << 2], add[maxn << 2];
    void pushUp(int rt){
        
        mx[rt] = max(mx[lson], mx[rson]);
    }
    void build(int l, int r, int rt){
 
        add[rt] = 0;
        if(l == r){
 
            mx[rt] = A - c[l];
            return;
        }
        int mid = gmid;
        build(l, mid, lson);
        build(mid + 1, r, rson);
        pushUp(rt);
    }
    void pushDown(int rt){
 
        if(add[rt]){
 
            add[lson] += add[rt], add[rson] += add[rt];
            mx[lson] += add[rt], mx[rson] += add[rt];
            add[rt] = 0;
        }
    }
    void update(int l, int r, int rt, int L, int R, ll val){
 
        if(l >= L && r <= R){
 
            add[rt] += val, mx[rt] += val;
            return;
        }
        int mid = gmid;
        pushDown(rt);
        if(L <= mid) update(l, mid, lson, L, R, val);
        if(R > mid) update(mid + 1, r, rson, L, R, val);
        pushUp(rt);
    }
    ll query(int l, int r, int rt, int L, int R){
 
        if(l >= L && r <= R) return mx[rt];
        int mid = gmid; ll ret = -oo;
        pushDown(rt);
        if(L <= mid) ret = max(ret, query(l, mid, lson, L, R));
        if(R > mid) ret = max(ret, query(mid + 1, r, rson, L, R));
        return ret;
    }
} trL, trR;
 
int fin(int x){
 
    return x == pre[x] ? x : pre[x] = fin(pre[x]);
}
 
void merge(int x, int y, ll D){
 
    int fx = fin(x), fy = fin(y);
    if(fx == fy) return;
    int l1 = li[fx], r1 = ri[fx];
    int l2 = li[fy], r2 = ri[fy];
    ll mxL = trL.query(1, n, 1, l1, r1);
    ll mxR = trR.query(1, n, 1, l2, r2);
    // cout << x << " " << y << " " << D << " " << mxL << " " << mxR << endl;
    ans = max(ans, mxL + mxR - D);
    trL.update(1, n, 1, l1, r1, sum[fy]);
    trR.update(1, n, 1, l2, r2, sum[fx]);
    li[fx] = min(li[fx], li[fy]);
    ri[fx] = max(ri[fx], ri[fy]);
    sum[fx] += sum[fy];
    pre[fy] = fx;
}
 
int main(){
 
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n >> A;
    for(int i = 1; i <= n; ++i) cin >> d[i] >> c[i];
    for(int i = 2; i <= n; ++i){
 
        tp[i] = {(d[i] - d[i - 1]) * (d[i] - d[i - 1]), i};
    }
    sort(tp + 2, tp + 1 + n);
    for(int i = 1; i <= n; ++i){
 
        pre[i] = li[i] = ri[i] = i;
        sum[i] = A - c[i];
        ans = max(ans, sum[i]);
    }
    trL.build(1, n, 1), trR.build(1, n, 1);
    for(int i = 2; i <= n; ++i){
 
        int y = tp[i].second, x = y - 1;
        merge(x, y, tp[i].first);
    }
    cout << ans << endl;
    return 0;
}

 
Monotone stack + RMQ:

#include<bits/stdc++.h>
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define sz(a) ((int)a.size())
#define pb push_back
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define gmid (l + r >> 1)
const int maxn = 3e5 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
 
ll d[maxn], c[maxn];
int li[maxn], ri[maxn], stk[maxn];
int n, A, top;

struct RMQ{

    ll mn[maxn][21], mx[maxn][21]; int lg[maxn];
    void build(){

        lg[1] = 0;
        for(int i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
        for(int i = 0; i <= n; ++i) mn[i][0] = mx[i][0] = c[i];
        int k = lg[n];
        for(int j = 1; j <= k; ++j){

            for(int i = 0; i <= n; ++i){

                if(i + (1 << j) - 1 > n) break;
                mn[i][j] = min(mn[i][j - 1], mn[i + (1 << (j - 1))][j - 1]);
                mx[i][j] = max(mx[i][j - 1], mx[i + (1 << (j - 1))][j - 1]);
            }
        }
    }
    ll query(int l, int r, int f){

        int k = lg[r - l + 1];
        if(f == 0) return min(mn[l][k], mn[r - (1 << k) + 1][k]);
        else return max(mx[l][k], mx[r - (1 << k) + 1][k]);
    }
} rmq;

int main(){
 
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n >> A;
    for(int i = 1; i <= n; ++i) cin >> d[i] >> c[i];
    for(int i = 1; i <= n; ++i){

        c[i] = A - c[i];
        c[i] += c[i - 1];
    }
    for(int i = n; i >= 2; --i){

        d[i] = (d[i] - d[i - 1]) * (d[i] - d[i - 1]);
    }
    top = 0, stk[0] = 1;
    for(int i = 2; i <= n; ++i){

        while(top && d[i] >= d[stk[top]]) --top;
        li[i] = stk[top] + 1, stk[++top] = i;
    }
    top = 0, stk[0] = n + 1;
    for(int i = n; i >= 2; --i){

        while(top && d[i] >= d[stk[top]]) --top;
        ri[i] = stk[top] - 1, stk[++top] = i;
    }
    for(int i = 2; i <= n; ++i) --li[i];
    ll ret = 0;
    rmq.build();
    for(int i = 1; i <= n; ++i) ret = max(ret, c[i] - c[i - 1]);
    for(int i = 2; i <= n; ++i){

        ll tmx = rmq.query(i, ri[i], 1);
        ll tmn = rmq.query(li[i] - 1, i - 2, 0);
        // cout << i << " " << li[i] << " " << ri[i] << " " << tmx << " " << tmn << endl;
        ret = max(ret, tmx - tmn - d[i]);
    }
    cout << ret << endl;
    return 0;
}