HDU 1506 dp求最大子矩阵 *

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本文介绍了如何通过动态规划（DP）算法来解决直方图中寻找最大矩形面积的问题。包括输入解析、计算每个矩形的左侧和右侧连续较大值的索引、以及最终通过这些信息计算最大面积的过程。提供了具体的代码实现和样例输入输出。

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14692 Accepted Submission(s): 4222

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

Source

University of Ulm Local Contest 2003

题意：让求出大的矩形面积
思路：

DP找出a[i]的左边和右边与自己连着的比自己大的数的长度,然后用这个长度乘以a[i],乘积最大的那个就是答案.

可以求出求a[i]之前的连续比他大的下标，求a[i]之后的连续比他大的下标
两下标相减就得长度了

#include<bits/stdc++.h>
using namespace std;
int l[100010],r[100100];
int a[100010];
int main()
{
    int n;
    while(~scanf("%d",&n),n)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            l[i]=r[i]=i;
        }
        for(int i=0;i<n;i++) //求a[i]之前的连续比他大的下标
        {
            while(l[i]>=1&&a[l[i]-1]>=a[i])
              l[i]=l[l[i]-1];
        }
        for(int i=n-1;i>=0;i--)//求a[i]之后的连续比他大的下标
        {
            while(r[i]<n-1&&a[r[i]+1]>=a[i])
                r[i]=r[r[i]+1];
        }
        long long Max=0,ans=0;
        for(int i=0;i<n;i++)
        {
            ans=(long long)(r[i]-l[i]+1)*a[i];  //左右下标相减
            if(ans>Max)
                Max=ans;
        }
        printf("%lld\n",Max);
    }
}