POJ 2559 Largest Rectangle in a Histogram (栈的运用）

最新推荐文章于 2021-05-12 14:52:29 发布

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最新推荐文章于 2021-05-12 14:52:29 发布

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原文链接：http://www.cnblogs.com/a249189046/p/7528325.html

本文介绍了一种求解直方图中最大矩形面积的高效算法。通过预处理找到每个柱状左侧和右侧最近的小于当前高度的位置，利用栈来实现这一过程。最终通过遍历所有柱状物并计算以每个柱状物为高的矩形面积，找出最大值。

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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.

分析：题目的数据很大，要求这个最大的面积，我们想到可能要预处理某些方面.

那么我们可以用L[i]表示从右往左找到的第一个j(j<i,且h[j]<h[i)]；

并且用R[i]表示从左往右找到的第一个j(j>i,且h[j]<h[i])

那么枚举所有i的 (R[i]-L[i]-1)*h[i]的最大值即为结果

用找L[i]举例，我们可以用栈进行处理,i的编号从1开始取

对于数据 7 2 1 4 5 1 3 3

L[1]=0 将1加入栈中 {1};

a[2]<a[1] 将1弹出栈中无元素故L[2]=0，再将2加入栈中{2}

a[3]>a[2] L[3]=2,将3加入栈中{2,3};

a[4]>a[3] L[4]=3,将4加入栈中{2,3,4};

a[5]<a[4] 将4弹出；

a[5]<a[3] 将3弹出;

a[5]<a[2] 将2弹出；此时栈为空 L[5]=0; 将5加入{5};

……

栈始终从栈顶到栈底，储存数列从大到小，从右到左的值，这样就总能找到在某个区域左边的第一个小于它的位置.

找R[i]的方法相似.

代码如下：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <map>
#include <algorithm>
#include <stack>
using namespace std;
typedef long long ll;
const int MAXN=1e5+10;
ll a[MAXN];
ll L[MAXN];
ll R[MAXN];
int main()
{

      int n;
      ll maxx;
      while(scanf("%d",&n)!=EOF&&n)
      {
          maxx=0;
             stack<ll>S;
            for(int i=1;i<=n;i++)
                scanf("%lld",&a[i]);
           L[1]=0;
           S.push(1);
            for(ll i=2;i<=n;i++)
           {
             while(!S.empty()&&a[S.top()]>=a[i])
              S.pop();
              if(S.empty())
                L[i]=0;
              else
              {
            //      cout<<" "<<S.top()<<endl;
                L[i]=S.top();
              }
            //  printf("%d\n",L[i]);
              S.push(i);
           }
           while(!S.empty())
            S.pop();
         R[n]=n+1;
           S.push(n);
           for(int i=n-1;i>=1;i--)
           {
             while(!S.empty()&&a[S.top()]>=a[i])
                S.pop();
             if(S.empty())
              R[i]=n+1;
             else
             {
                R[i]=S.top();
             }
              S.push(i);
           }
          for(int i=1;i<=n;i++)
           {
              maxx=max(maxx,(R[i]-L[i]-1)*a[i]);
           }
           printf("%lld\n",maxx);
      }

    return 0;
}

转载于:https://www.cnblogs.com/a249189046/p/7528325.html