HDU - 1506 Largest Rectangle in a Histogram（DP）

 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

题意：

给出直方图的模样，求出直方图中最大的矩形。

解题思路：

用dp[]数组记录，以当前位置的高度作为长方形的高时，最大的面积。即求最大长度

即求每个位置对应的左边界和右边界。左边界和右边界为i。

a[i]<=a[righ[i]+1]时，right[i]+1位置的右界，一定也满足i

a[i]<=a[lef[i]-1])时，left[i]-1位置的左界一定满足i

思路来源：

hdu 1506(dp求最大子矩阵)

---

#include
#include
using namespace std;
#define maxn 100005
long long a[maxn];
int lef[maxn],righ[maxn];
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        long long answer=0;
        for(int i=1;i<=n;i++)
        {
            //printf("**i=%d\n",i);
            scanf("%lld",&a[i]);
            lef[i]=i;
            righ[i]=i;
        }
        a[0]=-10000;
        a[n+1]=-10000;
        for(int i=n;i>=1;i--)
        {
            //printf("$$i=%d\n",i);
            while(a[i]<=a[righ[i]+1])//此时，right[i]+1位置的右界，一定也满足i
            {
                righ[i]=righ[righ[i]+1];
            }
        }
        for(int i=1;i<=n;i++)
        {
            //printf("##i=%d\n",i);
            while(a[i]<=a[lef[i]-1])//left[i]-1位置的左界一定满足i;
                lef[i]=lef[lef[i]-1];
            answer=max(answer,((long long)(righ[i]-lef[i]+1))*a[i]);
        }
        printf("%lld\n",answer);
    }
    return 0;
}