Maximum Subarray Revisited¶
Problem Description¶
Yun-Nung (Vivian) Chen's website
Concept¶
- Store the process of divide and conquer (just as segment tree)
- In each node representing [L, R], store 4 things:
- lmx = \max\limits_{L \leq r \leq R} \sum_{i=L}^{r} A_i
- rmx = \max\limits_{L \leq \ell \leq R} \sum_{i = \ell}^{R} A_i
- ans = \max\limits_{L \leq \ell \leq r \leq R} \sum_{i = \ell}^{r} A_i
- sum = \sum_{i=L}^{R} A_i
- lmx = \max\limits_{L \leq r \leq R} \sum_{i=L}^{r} A_i
- The "merge" step is as the code.
Code¶
#include <bits/stdc++.h>
#define IOS {ios::sync_with_stdio(false);cin.tie(nullptr);}
using namespace std;
using LL = int64_t;
using ULL = uint64_t;
using LD = long double;
inline int read_char();
inline int read_int(int &);
inline void write_int(LL);
constexpr int MAXN = 5e5+5;
struct Item {
LL lmx, rmx, sum, ans;
};
int N, A[MAXN];
Item seg[4 * MAXN];
inline int lch(int v) { return (v << 1); }
inline int rch(int v) { return (v << 1) | 1; }
inline void combine(Item &v, Item &a, Item &b) {
v.sum = a.sum + b.sum;
v.lmx = max(a.lmx, a.sum + b.lmx);
v.rmx = max(b.rmx, b.sum + a.rmx);
v.ans = max(a.rmx + b.lmx, max(a.ans, b.ans));
}
void build(int v, int tl, int tr) {
if ( tl == tr ) {
seg[v] = {A[tl], A[tl], A[tl], A[tl]};
} else {
int tm = (tl + tr) >> 1;
build(lch(v), tl, tm);
build(rch(v), tm+1, tr);
combine(seg[v], seg[lch(v)], seg[rch(v)]);
}
}
void pt_upd(int v, int tl, int tr, int pos, int new_v) {
if ( tl == tr ) {
seg[v] = {new_v, new_v, new_v, new_v};
} else {
int tm = (tl + tr) >> 1;
if ( pos <= tm ) pt_upd(lch(v), tl, tm, pos, new_v);
else pt_upd(rch(v), tm+1, tr, pos, new_v);
combine(seg[v], seg[lch(v)], seg[rch(v)]);
}
}
inline LL qry() {
return seg[1].ans < 0 ? 0 : seg[1].ans;
}
int main()
{
int Q;
read_int(N); read_int(Q);
for ( int i = 1; i <= N; i++)
read_int(A[i]);
build(1, 1, N);
write_int(qry());
while ( Q-- ) {
int pos, v; read_int(pos); read_int(v);
pt_upd(1, 1, N, pos, v);
write_int(qry());
}
return 0;
}
// input optimize
inline int read_char() {
constexpr int N = 1<<20;
static char buf[N];
static char *p = buf, *end = buf;
if ( p == end ) {
if ( (end = buf + fread(buf, 1, N, stdin)) == buf ) return EOF;
p = buf;
}
return *p++;
}
inline int read_int(int &x) {
static char c, neg;
while ( (c = read_char()) < '-' )
if ( c == EOF ) return 0;
neg = (c == '-') ? -1 : 1;
x = (neg == 1) ? c - '0' : 0;
while ( (c = read_char()) >= '0' )
x = (x<<3) + (x<<1) + (c-'0');
x *= neg;
return 1;
}
// output optimize
#define _pc putchar_unlocked
char _buf[128];
inline void write_int(LL x) {
if ( !x ) { _pc('0'); _pc('\n'); return; }
int t = 0;
while ( x ) {
_buf[t++] = x % 10 + '0';
x /= 10;
}
while ( t-- ) _pc(_buf[t]);
_pc('\n');
}
#undef _pc