C - Largest Rectangle in a Histogram

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, where 0<=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.


这道题是老师作为例题讲解的



设栈内的元素为一个二元组(x,y),x表示矩形的高度,y表示矩形的宽度。

若原始矩形高度分别为2,1,4,5,1,3,3

(1) 高度为2的元素进栈,当前栈为(2,1)

(2) 高度为1的元素准备进栈,但必须从栈顶开始删除高度大于或等于1的矩形,因为2已经不可能延续到当前矩形。删除(2,1)这个元素之后,更新最大矩形面积为2*1=2,然后把它的宽度1累加到当前高度为1的准备进栈的矩形,然后进栈,当前栈为(1,2)

(3) 高度为4的元素进栈,当前栈为(1,2)(4,1)

(4) 高度为5的元素进栈,当前栈为(1,2)(4,1) (5,1)

(5) 高度为1的元素准备进栈,删除(5,1)这个元素,更新最大矩形面积为5*1=5,把1累加到下一个元素,得到(4,2),删除(4,2),更新最大矩形面积为4*2=8,把2累加到下一个元素,得到(1,4),1*4=4<8,不必更新,删除(1,4),把4累加到当前准备进栈的元素然后进栈,当前栈为(1,5)

(6) 高度为3的元素进栈,当前栈为(1,5)(3,1)

(7) 高度为3的元素准备进栈,删除(3,1),不必更新,把1累加到当前准备进栈的元素然后进栈,当前栈为(1,5)(3,2)

(8) 把余下的元素逐个出栈,(3,2)出栈,不必更新,把2累加到下一个元素,当前栈为(1,7),(1,7)出栈,不必更新。栈空,结束。

最后的答案就是8。





#include 
#include 
#include 
using namespace std;

struct Rectangle
{
    int height;
    int weight;
};

stack rect;

int main()
{
    int height, n;
    long long ans, total, temp;

    while (scanf("%d", &n) != EOF && n)
    {
        ans = 0;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &height);
            temp= 0;
            while (!rect.empty() && rect.top().height >= height)
            {
                total = rect.top().height * (rect.top().weight + temp);
                if (total > ans) ans = total;
                temp+= rect.top().weight;
                rect.pop();
            }
            Rectangle data;
            data.height = height;
            data.weight = temp + 1;
            rect.push(data);
        }
        temp= 0;
        while (!rect.empty())
        {
            total = rect.top().height * (rect.top().weight + temp);
            if (total > ans) ans = total;
            temp+= rect.top().weight;
            rect.pop();
        }

        printf("%lld\n", ans);
    }
    return 0;
}


















你可能感兴趣的:(单调栈和单调队列)