POJ 2559 - Largest Rectangle in a Histogram（单调栈）

最新推荐文章于 2022-01-15 14:46:52 发布

原创最新推荐文章于 2022-01-15 14:46:52 发布 · 225 阅读

0 ·

CC 4.0 BY-SA版权

POJ 专栏收录该内容

35 篇文章

订阅专栏

本篇介绍了一种解决直方图中寻找最大矩形面积问题的方法，利用单调栈的思想来处理输入的矩形高度序列，通过不断调整栈内元素确保每次都能计算出可能的最大矩形面积。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Largest Rectangle in a Histogram
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 20194 Accepted: 6488
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:

这里写图片描述

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
Hint

Huge input, scanf is recommended.

题意：
找出最大矩形的面积。

解题思路：
对于一个矩形，如果他的前一个比他大，那么前一个就没有必要了，因为这等同于当前矩形的宽度加一。
所以用到单调栈的思想。
如果当前的这个小于或者等于栈顶，就让栈顶弹出，同时更新当前的向前长度和新栈顶的向后长度。同时将结果和弹出矩形的面积进行比较和更新（要保证栈不为空），最后将当前的矩形压栈。
当所有的矩形就压栈完毕，进行最后一次遍历。
对于栈顶的元素，都提出来，同时更新新栈顶的向后长度（因为栈是单调的），然后将结果与弹出的这个矩形面积进行比较更新。
这里要注意，因为输入量较大，不要使用cin，否则会超时。

AC代码：

#include<iostream>
#include<stdio.h>
#include<stack>
#include<algorithm>
using namespace std;
struct node
{
    long long height;
    long long pre;
    long long next;
};
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        stack<node>a;
        node tmp;
        scanf("%lld",&tmp.height);
        tmp.pre = 1;
        tmp.next = 1;
        a.push(tmp);
        n--;
        long long res = 0;
        while(n--)
        {
            scanf("%lld",&tmp.height);
            tmp.pre = 1;
            tmp.next = 1;
            while(!a.empty()&&a.top().height>=tmp.height)
            {
                node tmp1 = a.top();a.pop();
                res = max(res,tmp1.height*(tmp1.pre+tmp1.next-1));
                if(!a.empty())  a.top().next += tmp1.next;
                tmp.pre += tmp1.pre;
            }
            a.push(tmp);
        }
        while(!a.empty())
        {
            node tmp1 = a.top();a.pop();
            if(!a.empty())  a.top().next += tmp1.next;
            res = max(res,tmp1.height*(tmp1.pre+tmp1.next-1));
        }
        printf("%lld\n",res);
    }
    return 0;
}