Monotone Queue Water Brush ads

"Problem description"
Recently, Afy decided to give Toj print ads, billboards are painted in the city's buildings, the city has a close to the N buildings.
Afy decided to place a billboard on top of a rectangle as large as possible. We assume that each building has a height,
From left to right give the height of each building h1,h2 ... HN, and 0

Requires the maximum area of the output billboard.

"Input File"
The first line in the input file ad.in is a number n (n<= 400,000)
The second line is the number of N, which indicates the height of each building h1,h2 ... HN, and 0"Output File"
A total of one row in the output file Ad.out, representing the maximum area of the billboard.
"Input Sample"
6
5 8 4 4 8 4
"Output Example"

24

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring>using namespace STD; #define MAXN 400000 + 10int L[MAXN], r[maxn];int pos[maxn];int a[maxn];int n;void get_left () {int head = 1, tail =    0;        for (int i = 1; I <= N, i++) {while (Head <= tail && a[pos[tail]] >= a[i]) tail--;        L[i] = I-pos[tail]-1;    Pos[++tail] = i;    }}void get_right () {int head = 1, tail = 0;    Pos[tail] = n + 1;        for (int i = n, i >= 1; i--) {while (Head <= tail && a[pos[tail]] >= a[i]) tail--;        R[i] = Pos[tail]-i-1;    Pos[++tail] = i;        }}int Main () {while (~SCANF ("%d", &n) {memset (l, 0, sizeof L);        memset (r, 0, sizeof R);        for (int i = 1; I <= n; i++) scanf ("%d", A + i);        Get_left ();        Get_right ();        Long Long ans =-1; for (int i = 1; I <= n; i++) ans = max (ans, (long long) l[i] + r[i] + 1) * A[i]);    printf ("%i64d\n", ans); } return 0;} /*65 8 4 4 8 4*/

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Monotone Queue Water Brush ads