# NC50965 Largest Rectangle in a Histogram

## 题目

A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles:

Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.

The input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that $$1 \leq n \leq 100000$$ . Then follow n integers $$h1\dots hn$$, where $$0 \leq h_i \leq 1000000000$$. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case.

For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line.

7 2 1 4 5 1 3 3
4 1000 1000 1000 1000
0


8
4000


Huge input, scanf is recommended.

## 题解

### 代码

#include <bits/stdc++.h>

using namespace std;

int h[100007];
int l[100007], r[100007];
///最大矩形高度肯定是某个矩形高度
///对于一个矩形，水平扩展距离取决于第一个比他小的，两边都是
///于是对每个矩形，用单调递增栈获得他左侧/右侧第一个比它小的矩形位置，就能知道左侧/右侧扩展距离
int main() {
std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int n;
while (cin >> n, n) {
for (int i = 0;i < n;i++) cin >> h[i];
stack<int> s1;
for (int i = 0;i < n;i++) {
while (!s1.empty() && h[s1.top()] >= h[i]) s1.pop();
l[i] = s1.empty() ? 0 : s1.top() + 1;///左侧大于等于的第一个位置
s1.push(i);
}
stack<int> s2;
for (int i = n - 1;i >= 0;i--) {
while (!s2.empty() && h[s2.top()] >= h[i]) s2.pop();///一定是大于等于，于是栈就是严格递减栈，元素是最靠右的
r[i] = s2.empty() ? n - 1 : s2.top() - 1;///右侧大于等于的最后一个位置
s2.push(i);
}
long long ans = 0;
for (int i = 0;i < n;i++)
ans = max(ans, (r[i] - l[i] + 1LL) * h[i]);
cout << ans << '\n';
}
return 0;
}