# Hdu1506 (DP)

Source: Internet
Author: User
Largest rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 11219 accepted submission (s): 3074

Problem descriptiona 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.

Inputthe 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. 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.

Outputfor 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 input7 2 1 4 5 1 3 34 1000 1000 1000

Sample output84000 dynamic plan to find out the length of the number on the left and right of a [I] connected to itself, and then multiply it by a [I], the one with the largest product is the ID that can be reached on the left using L [I], and the ID that can be reached on the right using R [I, the area is (R [I]-l [I] + 1) * A [I]
` 1 #include<stdio.h> 2 #include<string.h> 3 __int64 a[100010],L[100010],r[100010],max; 4 int main() 5 { 6     int i,j,n,t; 7     while(~scanf("%d",&n)&&n) 8     { 9         for(i=1; i<=n; i++)10             scanf("%I64d",&a[i]);11         L[1]=1;12         r[n]=n;13         for (i=2; i<=n; i++)14         {15             t=i;16             while (t>1 && a[i]<=a[t-1])17                 t=L[t-1];18             L[i]=t;19         }20         for (i=n-1; i>=1; i--)21         {22             t=i;23             while (t<n && a[i]<=a[t+1])24                 t=r[t+1];25             r[i]=t;26         }27         max=0;28         for(i=1; i<=n; i++)29         {30             if((r[i]-L[i]+1)*a[i]>max)31                 max=(r[i]-L[i]+1)*a[i];32         }33         printf("%I64d\n",max);34     }35     return 0;36 }`
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