「SP1043」GSS1 - Can you answer these queries I
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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)
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