POJ 2559 Largest Rectangle in a Histogram【解法二】

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.

解法一见【这里】。
单调栈。
对于每个点维护左边和右边低于他的第一个位置，记为l[i]，r[i]。
则答案为max{h[i]*(r[i]-l[i]-1)|1<=i<=n}。
接下来考虑如何求r数组和l数组。
以求r数组为例。
从左往右扫描，维护高度不减的栈。
弹栈的时候，所有被弹出去的点的r即为当前的i。
求l同理，从右往左扫描即可。
为了方便，可以令h[0]=h[n+1]=-1。

#include
#include
long long max(long long a,long long b)
{
    return a>b?a:b;
}
int l[1000010],r[1000010],h[1000010],s[1000010];
int main()
{
    int i,j,k,m,n,p,q,x,y,z,t;
    long long ans;
    while (scanf("%d",&n)&&n)
    {
        ans=0;
        for (i=1;i<=n;i++)
          scanf("%d",&h[i]);
        h[0]=h[n+1]=-1;
        t=0;
        for (i=1;i<=n+1;i++)
        {
            while (t&&h[s[t]]>h[i])
            {
                r[s[t]]=i;
                t--;
            }
            s[++t]=i; 
        }
        t=0;
        for (i=n;i>=0;i--)
        {
            while (t&&h[s[t]]>h[i])
            {
                l[s[t]]=i;
                t--;
            }
            s[++t]=i; 
        }
        for (i=1;i<=n;i++)
          ans=max(ans,(long long)h[i]*(r[i]-l[i]-1));
        printf("%lld\n",ans);
    }
}