hdu 1506 Largest Rectangle in a Histogram

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11248    Accepted Submission(s): 3088

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

题目让求一个面积最大的矩形。

思路：对于每个单位矩形高度为h[x],从当前位置向两边找出高度小于它的位置，

用数组l[i],r[i],分别记录左右两边高度大于等于h[i]的位置。

#include
#include
#include
#include
#include
using namespace std;
#define N 100005
#define ll __int64
int h[N],l[N],r[N];
int main()
{
    int i,j,n;
    while(scanf("%d",&n),n)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&h[i]);
            l[i]=r[i]=i;
        }
        h[0]=h[n+1]=-1;
        for(i=1;i<=n;i++)
        {
            j=i-1;
            while(h[j]>=h[i]) // 找到最左边高度小于当前高度的坐标
            {
                j=l[j]-1;     //l[j]肯定满足条件，则减一
            }
            l[i]=j+1;        //j位置小于h[i],则j+1位置满足条件
        }
        for(i=n;i>0;i--)   //同上
        {
            j=i+1;
            while(h[j]>=h[i])
            {
                j=r[j]+1;
            }
            r[i]=j-1;
        }
        ll ans=0;
        for(i=1;i<=n;i++)
            ans=max(ans,(ll)(r[i]-l[i]+1)*h[i]);  //结果会超出整数int范围
        printf("%I64d\n",ans);
    }
    return 0;
}