【HDU 1506】Largest Rectangle in a Histogram（DP）

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

题目大意

求直方图的最大面积，也就是在直方图中找到一个最大的矩形。

思路

定义dp[i]表示柱状图中i位置所能构成最大矩形的面积,arr[i]表示i位置柱状图的高度，l[i]表示柱状图中i位置往左搜索其能构成最大矩形的左边界的位置，r[i]表示柱状图中i位置往右搜索其能构成最大矩形的右边界的位置。那么对于dp[i]有如下公式：

dp[i]=(r[i]-l[i])*arr[i];

那么我们只需要求出每个点的左右边界就可以求出每个点所能构成最大矩形的面积，最后找出最大的一个输出即可。注意数据范围，数组要开long long。

代码

#include 
#include 
#include 
#include 
#define LL long long

using namespace std;
const int maxn=100000+5;

LL arr[maxn],l[maxn],r[maxn],ans;

int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&arr[i]);
        }
        l[1]=1;
        r[n]=n;
        ans=-1;
        for(int i=2;i<=n;i++)//寻找左边界
        {
            int tmp=i;
            while(tmp>1 && arr[i]<=arr[tmp-1]) tmp=l[tmp-1];
            l[i]=tmp;
        }
        for(int i=n-1;i>=1;i--)//寻找右边界
        {
            int tmp=i;
            while(tmp1]) tmp=r[tmp+1];
            r[i]=tmp;
        }
        for(int i=1;i<=n;i++)//寻找最大面积
        {
            if((r[i]-l[i]+1)*arr[i]>ans) ans=(r[i]-l[i]+1)*arr[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}