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

Posted 2022-12-30 SSL_LKJ

Largest Rectangle in a Histogram

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.

Hint

Huge input, scanf is recommended.

输入样例

7 2 1 4 5 1 3 3
4 1000 1000 1000 1000
0

输出样例

8
4000

解题思路

单调递增栈模板

AC代码

#include<iostream>
#include<cstdio>
using namespace std;
long long n,o,s,tail,a[100005];
struct node

	long long x,y;
f[100005];
int main()

	scanf("%lld",&n);
    while(n!=0)
    
     	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
     	s=tail=0;
     	for(int i=1;i<=n;i++)
     	
     	 	o=0;
     	 	while(f[tail-1].x>=a[i]&&tail>0) //单调栈
     	 	
     	 	   s=max(s,f[tail-1].x*(f[tail-1].y+o));
     	 	   o+=f[tail-1].y; 
     	 	   tail--;
     	 	
     	 	f[tail].x=a[i];
     	 	f[tail].y=o+1;
     	 	tail++;
     	
     	o=0;
     	while(tail>0)
     	
     	    s=max(s,f[tail-1].x*(f[tail-1].y+o));
     	 	o+=f[tail-1].y;
     	 	tail--;
     	
     	printf("%lld\\n",s);
     	scanf("%lld",&n);
    

谢谢