Largest Rectangle in a Histogram

　　　　　　　　　　　　　　　　Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　　　　　　Total Submission(s): 13037    Accepted Submission(s): 3680


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
 

题意：求木板连在一起的最大矩形面积

分析：一个木板左边连续比他高的最后一块木板的坐标为l[i]，右边连续比他高的最后一块木板的坐标为r[i]，面积则为(r[i]-l[i]+1)*a[i],但按原始的思路会超时，需要一些优化，把一些坐标存起来，具体操作看代码

代码：

#include "stack"
#include "stdio.h"
#include "iostream"
#include "math.h"
#include "string.h"
#include "algorithm"
#include "queue"
#define LL long long
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
int i,j,r[100005],l[100005],n,k;
LL a[100005],mas;
int main()
{
　　while(scanf("%d",&n)!=EOF)
　　{
　　　　if(n==0)
　　　　　　break;
　　　　for(i=1;i<=n;i++)
　　　　　　scanf("%lld",&a[i]);
　　　　l[1]=1;
　　　　for(i=2;i<=n;i++)
　　　　{
　　　　　　k=i;
　　　　　　while(k>1&&a[i]<=a[k-1])
　　　　　　　　k=l[k-1];
　　　　　　l[i]=k;
　　　　}
　　　　r[n]=n;
　　　　for(i=n-1;i>=1;i--)
　　　　{
　　　　　　k=i;
　　　　　　while(k<n&&a[i]<=a[k+1])
　　　　　　　　k=r[k+1];
　　　　　　r[i]=k;
　　　　}
　　　　mas=-1;
　　　　for(i=1;i<=n;i++)
　　　　{
　　　　　　if((r[i]-l[i]+1)*a[i]>mas)
　　　　　　　　mas=(r[i]-l[i]+1)*a[i];
　　　　}
　　　　printf("%lld\n",mas);
　　}
return 0;
}