POJ2559 Largest Rectangle in a Histogram

Largest Rectangle in a Histogram

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26987		Accepted: 8727

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

 

 

****这道题是单调栈的题stact<intfloatdouble> st是建造一个栈，st是这个栈的名字，st.push(i)表示把i放进这个栈里面，st.top()栈顶元素，st.pop（）删除栈顶元素，st.empty()判断这个栈是否为空，st.size()返回这个栈的大小。

复杂度o(n)

应用：

1.最基础的应用就是给定一组数，针对每个数，寻找它和它右边第一个比它大的数之间有多少个数。

2.给定一序列，寻找某一子序列，使得子序列中的最小值乘以子序列的长度最大。

3.给定一序列，寻找某一子序列，使得子序列中的最小值乘以子序列所有元素和最大。

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
using namespace std;
int main()
{
    int i,n,top;//top指向栈顶 
    stack<int> st;//栈用于保存矩形的编号，即位置 
    long long int tmp,ans,a[100010];//tmp为临时变量，记录面积的值，ans为结果，记录最大面积值 
    while(scanf("%d",&n) && n)
    {
        for(i = 1;i <= n;i++)
        {
            scanf("%lld",&a[i]);
        }
        ans = 0;
        a[n + 1] = -1; //最后一个元素设为最小值，以最后清空栈 
        for(i = 1;i <= n + 1;i++)
        {
            if(st.empty() || a[i] >= a[st.top()])
            { //如果栈为空或入栈元素大于等于栈顶元素则入栈 
                st.push(i);
            }
            else
            {
                while(!st.empty() && a[i] < a[st.top()])
                {//如果栈非空且入栈元素小于栈顶元素，则将栈顶元素出栈 
                    top = st.top();
                    st.pop();
                    tmp = (i - top) * a[top];//在出栈过程中计算面积值 
                    if(tmp > ans)//更新面积最大值 
                    {
                        ans = tmp;
                    }
                }
                st.push(top);//将这个值更新到最新的位置 
                a[top] = a[i];
            }
        }
        printf("%lld
",ans);
     } 
     return 0;
 }