[poj2559] 最大矩形面积（单调栈、单调队列）

 

 

Largest Rectangle in a Histogram

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 32762		Accepted: 10667

                                                                       

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

 

//老师讲了单调栈然鹅还是用了单调队列orz不过单调队列真好用_(:3」/)_

#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int n, top, a, ans=INT_MIN;
struct polygon{
    int height, pos;
}ss[N];
int main()
{        
         cin>>n; 
         for (int i = 1; i <= n; i++){          
          scanf("%d", &a);            
          while (top > 0 && a <= ss[top].height){
           int x = ( (i - 1) - ss[top - 1].pos)* ss[top].height;              
           if (ans < x ) ans = x;   
           top--;             
           }            
          top++;            
          ss[top].height = a;            
          ss[top].pos = i;        
        }  
              
        while (top > 0){ 
         int y = ( (n - ss[top -1].pos))*ss[top].height;           
         if (ans < y) ans = y;
             top--;        
        } 
        cout << ans; 
}