hdu 1506 Largest Rectangle in a Histogram（DP）

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11389    Accepted Submission(s): 3133

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

 

就是一段区间，取这段区间的最小值，乘以区间长度，找最大值，思路就是找左边比它大的，右边比它大的，枚举一下就可以，暴力找肯定超时，用dpl[i]计一下左边不小于的，如果左边第一个不小于它，则左边dpl[i-1]个都不小于它，则这dpl[i-1]个就课以跳过，再往后找，直到找到比它小的。同理右边。

代码：

#include 
#include 
#include 
#include 
using namespace std;
const int maxn=100000+100;
int a[maxn];
int dpl[maxn],dpr[maxn];
int main()
{
     int n;
     while(~scanf("%d",&n))
     {
         if(n==0)
         break;
         for(int i=1;i<=n;i++)
         {
             scanf("%d",&a[i]);
             int s=i-1;
             dpl[i]=1;
             while(a[s]>=a[i]&&s>0)//左边比它大的个数+1
             {
                 dpl[i]+=dpl[s];
                 s-=dpl[s];
             }
         }
         for(int i=n;i>=1;i--)
         {
             int s=i+1;
             dpr[i]=1;
             while(a[s]>=a[i]&&s<=n)//右边比它大的个数+1
             {
                 dpr[i]+=dpr[s];
                 s+=dpr[s];
             }
         }
         long long max=0;
         long long ans;
         for(int i=1;i<=n;i++)
         {
             ans=(long long)a[i]*(dpl[i]+dpr[i]-1);
             if(ans>max)
             max=ans;
         }
         printf("%I64d\n",max);
     }
    return 0;
}