Largest Rectangle in a Histogram (单调栈)-优快云博客

本文介绍了一种使用单调栈求解直方图中最大矩形面积的问题，通过维护一个递增栈并计算每个柱状图可能形成的最大矩形面积来解决。文章提供了完整的代码实现，并解释了核心算法思想。

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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.

Input

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<=n<=100000. Then follow n integers h1,...,hn, where0<=hi<=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.

Output

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.

Sample Input

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

Sample Output

8
4000

Hint

Huge input, scanf is recommended.

题意：求图上最大的矩形面积

解：用单调栈存数字，且是升序存储，每当栈内数字要退栈时计算那个数字的向左向右延长的矩阵，每次退栈时，把被退栈的长度加在刚入栈的元素长度上，因为被退栈的元素大于刚入栈元素。所以计算的那个矩阵，左边长度就是它的长度，因为左边比他长的已经加在他长度上了，右边长度是当前（i - 栈元素位置-1 ）

#include<map>
#include<vector>
#include<iostream>
#include<queue>
#include<deque>
#include<math.h>
#include<algorithm>
#include<stdio.h>
#include<string>
#include<string.h>
#define ll long long
const int maxn=1e7+10;
using namespace std;
int a[1100000];
struct node
{
    ll h,l;
}e[1100000];
int q[1100000];
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&e[i].h);
            e[i].l=1;
        }
        int l=0,r=1;
        ll ans=0;
        q[0]=1;
        for(int i=2;i<=n;i++)
        {
            while(l<r&&e[q[r-1]].h>=e[i].h)
            {
               ans=max(ans,e[q[r-1]].h*(i-q[r-1]-1+e[q[r-1]].l));
               e[i].l+=e[q[r-1]].l;
               r--;
            }
            q[r++]=i;
        }
        while(l<r)
        {
            ans=max(ans,e[q[r-1]].h*e[q[r-1]].l);
           if(r>=2) e[q[r-2]].l+=e[q[r-1]].l;
            r--;
        }
        printf("%lld\n",ans);
    }
}