POJ2559 Largest Rectangle in a Histogram —— 单调栈

题目链接：http://poj.org/problem?id=2559

Largest Rectangle in a Histogram

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24259		Accepted: 7844

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,...,h_n, where 0<=h_i<=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

Hint

Huge input, scanf is recommended.

Source

Ulm Local 2003

题解：

一开始想到的是;从当前元素，向前向后延伸，计算最多能延伸多长。O（n^2）超时~~~~~

然后看了网上的题解：单调栈。

如果当前元素大于栈顶元素，则入栈；否则，栈顶元素出栈，直到栈顶元素小于当前元素或栈为空为止。期间更新出栈元素能达到的宽度，因为上一次出栈的元素的高度必定大于这次出栈元素的高度，所以他对这次出栈元素的宽度有贡献，并更新ans。出栈完毕后，再将已出栈元素的总宽度叠加到当前元素中，因为这些元素的高度都大于等于当前元素的高度，所以这些元素对当前元素的宽度都有贡献。 最后栈中可能还会有元素，把他“倒”出来再统计就是了。

代码如下：

#include<cstdio>//poj2559 单调栈 此栈从栈底到栈顶为严格递增
#include<cstring>
#define MAX(a,b) (a>b?a:b)
#define LL long long
#define mod 1000000007

using namespace std;

struct Stack
{
    int he;
    int len;
}s[100010];

int main()
{
    int n,h;
    while(scanf("%d",&n) && n)
    {
        LL ans = -1;
        int top = 0;//从1开始存放，top指向栈顶元素
        for(int i = 1; i<=n; i++)
        {
            scanf("%d",&h);
            int len = 0;
            //当栈顶元素的高度大于等于当前将入栈的元素的高度时，出栈，并更新其之前元素的宽度
            while(top>0 && h<=s[top].he)
            {
                len += s[top].len;
                ans = MAX(ans,1LL*s[top].he*len);
                top--;
            }
            //更新入栈元素的宽度
            s[++top].len = len + 1;
            s[top].he = h;
        }

        int len = 0;
        while(top>0)//将栈内剩余的元素“倒”出来统计
        {
            len += s[top].len;
            ans = MAX(ans,1LL*len*s[top].he);
            top--;
        }

        printf("%lld
",ans);
    }
    return 0;
}