86 lines
2.4 KiB
C++
86 lines
2.4 KiB
C++
#include <cstdio>
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#include <cstring>
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#include <algorithm>
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#include <iostream>
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using namespace std;
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const int N = 10010;
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int a[N];
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struct Node {
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int l, r;
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int sum, tmax, lmax, rmax;
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} tr[N << 2];
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void calc(Node &u, Node &l, Node &r) {
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u.sum = l.sum + r.sum; // 区间和
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u.tmax = max({l.tmax, r.tmax, l.rmax + r.lmax}); // 子区间最大值
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u.lmax = max(l.lmax, l.sum + r.lmax); // 左前缀最大值
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u.rmax = max(r.rmax, r.sum + l.rmax); // 右后缀最大值
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}
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void pushup(int u) {
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calc(tr[u], tr[u << 1], tr[u << 1 | 1]);
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}
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void build(int u, int l, int r) {
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tr[u] = {l, r, 0, 0, 0, 0};
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if (l == r) {
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tr[u].sum = tr[u].tmax = tr[u].lmax = tr[u].rmax = a[l];
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return;
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}
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int mid = (tr[u].l + tr[u].r) >> 1;
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build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
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pushup(u);
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}
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Node query(int u, int l, int r) {
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if (l > tr[u].r || r < tr[u].l) return {}; // 递归出口，不在我管辖范围内的情况，返回0
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if (l <= tr[u].l && tr[u].r <= r) return tr[u];
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int mid = (tr[u].l + tr[u].r) >> 1;
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if (r <= mid) return query(u << 1, l, r);
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if (l > mid) return query(u << 1 | 1, l, r);
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Node a = query(u << 1, l, r), b = query(u << 1 | 1, l, r), c;
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calc(c, a, b);
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return c;
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}
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int main() {
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// 加快读入
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ios::sync_with_stdio(false), cin.tie(0);
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int T, n, m;
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cin >> T;
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while (T--) {
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cin >> n;
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for (int i = 1; i <= n; i++) cin >> a[i];
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build(1, 1, n);
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cin >> m;
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while (m--) {
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int x1, y1, x2, y2, res;
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Node a, b, c;
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cin >> x1 >> y1 >> x2 >> y2;
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if (y1 < x2) { // 这里不能取等 不然边界会被算2次
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a = query(1, x1, y1);
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b = query(1, x2, y2);
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c = query(1, y1 + 1, x2 - 1);
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res = a.rmax + c.sum + b.lmax;
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} else {
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res = query(1, x2, y1).tmax; // 最大子序列和出现在交集中
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a = query(1, x1, x2 - 1);
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b = query(1, x2, y2);
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res = max(res, a.rmax + b.lmax);
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a = query(1, x1, y1);
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b = query(1, y1 + 1, y2);
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res = max(res, a.rmax + b.lmax);
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}
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printf("%d\n", res);
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}
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}
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return 0;
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} |