POJ 2559 Largest Rectangle in a Histogram【解法一】

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.

解法二见【这里】。
单调栈模板题。
维护一个高度单调递增的栈，对于每个元素弹出栈顶并利用它更新答案，直到序列递增。
弹出栈顶的时候维护宽度，用来更新答案以及新元素入栈。
最后把栈弹干净。

#include<cstdio>#include<cstring>long long max(long long a,long long b){    return a>b?a:b;}struct rec{    int h,l;}r1,r2,sta[1000010];int hei[1000010];int main(){    int i,j,k,m,n,p,q,top;    long long x,y,z,ans;    while (scanf("%d",&n)&&n)    {        ans=top=0;        for (i=1;i<=n;i++) scanf("%d",&hei[i]);        for (i=1;i<=n;i++)        {            x=0;            while (top&&sta[top].h>hei[i])            {                x+=sta[top].l;                ans=max(ans,(long long)x*sta[top].h);                top--;            }            r1.h=hei[i];            r1.l=x+1;            sta[++top]=r1;        }        x=0;        while (top)        {            x+=sta[top].l;            ans=max(ans,(long long)x*sta[top].h);            top--;        }        printf("%lld\n",ans);    }}