Largest Rectangle in a Histogram

最新推荐文章于 2022-01-15 14:46:52 发布

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本文介绍了一种高效算法，用于解决计算直方图中最大矩形面积的问题。该算法通过预处理每个点能够向两边延伸的距离来寻找最大的矩形面积，并提供了一个C++实现示例。

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.

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

思路：

预处理出每一个点能够向两边延伸的距离，向左找到第一个比当前点高度小的下标记为i，向右找到第一个比当前点高度小的下标记为j，那么此时以这点高度的最大值是f[i]*(j-i+1); 然后用for循环扫一遍求出最大值即可。

代码：

#include<stdio.h>
const int N = 100010;
__int64 f[N],l[N],r[N],sum;
int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF&&n)
    {
        sum=0;
        for(i=1;i<=n;i++)
    {
        scanf("%I64d",&f[i]);
        l[i]=r[i]=i;
    }
    f[0]=f[n+1]=-1;
    for(i=1;i<=n;i++)
    {
        while(f[i]<=f[l[i]-1])
            l[i]=l[l[i]-1];
    }
    for(i=n;i>=1;i--)
    {
        while(f[i]<=f[r[i]+1])
            r[i]=r[r[i]+1];
    }
    for(i=1;i<=n;i++)
        if(f[i]*(r[i]-l[i]+1)>sum)
        sum=f[i]*(r[i]-l[i]+1);
    printf("%I64d\n",sum);
    }
    return 0;
}