Largest Rectangle in a Histogram（最大长方形）

最新推荐文章于 2019-08-30 13:25:32 发布

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最新推荐文章于 2019-08-30 13:25:32 发布

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博客探讨了如何在直方图中找到最大的矩形区域，即寻找具有最大面积的矩形问题，这个问题在计算机科学中常用于数据处理和算法设计。

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Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17827 Accepted Submission(s): 5330

Problem Description

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

题解：题目大意为给你一个n,表示n个矩形，然后n个矩形的高，求在这些矩形中所能找到的最大矩形面积……
由于n为10^5,并且时间限制为1秒，所以只能一层for，开三个数组x,y,z。x存矩形的高度，y存x[i]左边的矩形高度不小于x[i]的下标，z存x[i]右边的矩形高度不小于x[i]的下标，最后遍历求面积最大值就行了……

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define inf 1000000000
long long   q1[100005],l[100010],r[100010];   // q1存高度，l存左边的坐标，r存右边的坐标
int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        for(int i=1; i<=n; i++)
        {
        scanf("%lld",&q1[i]);
        }
        q1[0]=-1;                           // 处理下边界
        q1[n+1]=-1;
        for(int i=1; i<=n; i++)             // 赋初值
        {
            l[i]=i;
            r[i]=i;
        }
        for(int i=1; i<=n; i++)
        {
            int e=i-1;
            while(q1[i]<=q1[e])
            {
                l[i]=l[l[e]];
                e=l[i]-1;
            }
        }
        for(int i=n; i>=1; i--)
        {
            int e=i+1;
            while(q1[i]<=q1[e])
            {
                r[i]=r[r[e]];
                e=r[i]+1;
            }
        }
        long long ans=0;
        for(int i=1; i<=n; i++)             // 遍历找最大值
        {
            ans=max(ans,(long long )q1[i]*(r[i]-l[i]+1));
        }
        printf("%lld\n",ans);
    }

}