「SP1043」GSS1

传送门
Luogu

解题思路

这题就是 GSS3 的一个退化版，不带修改操作的区间最大子段和，没什么好讲的。

细节注意事项

• 咕咕咕

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#define rg register
using namespace std;
template < typename T > inline void read(T& s) {
 	s = 0; int f = 0; char c = getchar();
 	while (!isdigit(c)) f |= (c == '-'), c = getchar();
 	while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
 	s = f ? -s : s;
}

const int _ = 50000 + 10;

int n, m, a[_];
struct node{ int sum, L, R, mx; }t[_ << 2];

inline int lc(int p) { return p << 1; }

inline int rc(int p) { return p << 1 | 1; }

inline void pushup(int p) {
	t[p].sum = t[lc(p)].sum + t[rc(p)].sum;
	t[p].L = max(t[lc(p)].L, t[lc(p)].sum + t[rc(p)].L);
	t[p].R = max(t[rc(p)].R, t[rc(p)].sum + t[lc(p)].R);
	t[p].mx = max(t[lc(p)].R + t[rc(p)].L, max(t[lc(p)].mx, t[rc(p)].mx));
}

inline void build(int p = 1, int l = 1, int r = n) {
	if (l == r) { t[p] = (node) { a[l], a[l], a[l], a[l] }; return; }
	int mid = (l + r) >> 1;
	build(lc(p), l, mid), build(rc(p), mid + 1, r), pushup(p);
}

inline node query(int ql, int qr, int p = 1, int l = 1, int r = n) {
	if (ql <= l && r <= qr) return t[p];
	int mid = (l + r) >> 1;
	if (ql > mid) return query(ql, qr, rc(p), mid + 1, r);
	if (qr <= mid) return query(ql, qr, lc(p), l, mid);
	node ls = query(ql, mid, lc(p), l, mid);
	node rs = query(mid + 1, qr, rc(p), mid + 1, r);
	node res;
	res.sum = ls.sum + rs.sum;
	res.L = max(ls.L, ls.sum + rs.L);
	res.R = max(rs.R, rs.sum + ls.R);
	res.mx = max(ls.R + rs.L, max(ls.mx, rs.mx));
	return res;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.in", "r", stdin);
#endif
	read(n);
	for (rg int i = 1; i <= n; ++i) read(a[i]);
	build();
	read(m);
	for (int ql, qr; m--; )
		read(ql), read(qr), printf("%d
", query(ql, qr).mx);
	return 0;
}

完结撒花 (qwq)