ZOJ 1985 Largest Rectangle in a Histogram（刷广告）2010辽宁省赛

Largest Rectangle in a Histogram

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21204		Accepted: 6831

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 nintegers 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

 

 

 

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<stack>
using namespace std;

struct node
{
    int h,id;
    node(int a,int b){ h=a; id=b; }  //h存高度，id存第几块
};
int n;
stack<node> Q;
long long a[100005],l[100005],r[100005];
//在a[i]的高度下，能达到的左边界l[i]，能达到的右边界r[i]

int main()
{
    while(~scanf("%d",&n))
    {
        if(n==0) break;
       for(int i=1;i<=n;i++) scanf("%d",&a[i]);
       while(!Q.empty()) Q.pop();     //清空栈Q
       memset(l,0,sizeof(l));
       memset(r,0,sizeof(r));
      //求l[]数组，O(n)复杂度找左边界
       for(int i=1;i<=n;i++)
       {
         if (Q.empty()) {l[i]=1; Q.push(node(a[i],i)); continue; }
         node u=Q.top();
         if (u.h<a[i]) {l[i]=i; Q.push(node(a[i],i)); continue;}
         while(u.h>=a[i]) //如果当前高度a[i]，比栈顶元素低，则栈顶元素能到达的左边界也能到达。
         {
             l[i]=l[u.id];
             Q.pop();
             if (Q.empty()) break;
             u=Q.top();
         }
         Q.push(node(a[i],i) );
       }

       //求r[]数组，扫右边界
       while(!Q.empty()) Q.pop();
       for(int i=n;i>=1;i--)
       {
         if (Q.empty()) {r[i]=i; Q.push(node(a[i],i)); continue; }
         node u=Q.top();
         if (u.h<a[i]) {r[i]=i; Q.push(node(a[i],i));continue;}
         while(u.h>=a[i])
         {
             r[i]=r[u.id];
             Q.pop();
             if (Q.empty()) break;
             u=Q.top();
         }
         Q.push(node(a[i],i) );
       }

       long long sum=0;
       for(int i=1;i<=n;i++)
           sum=max(sum,a[i]*(r[i]-l[i]+1));
       printf("%lld
",sum);
    }
    return 0;
}