HDU-1506-Largest Rectangle in a Histogram——算法笔记

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1506

Problem Description：

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.

Sample Input：

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

Sample Output：

8
4000

主要思路：由于每个最小单位长的矩形的高知道（即输入的数组值），那么只用计算出该位置所能向最左、最右扩展的下标的最大值就行了。最后遍历每个数据，把最右、最左值相减得到最大矩形的长，乘以对应的高就是当前数据所得到的最大矩形面积，再找出最大的那个面积就行了。对着代码动手计算一遍就明白了。

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;

#define ll long long
#define clean(arrays) memset(arrays, 0, sizeof(arrays))

ll a[100005], lefts[100005], rights[100005];        // lefts 和 rights 数组都表示的是下标
ll max_area;

int main()
{
    int n, t;
    while (cin>>n, n != 0)
    {
        clean(a);   clean(lefts);    clean(rights);
        for (int i = 1; i <= n; i++)
            cin>>a[i];

        lefts[1] = 1;   // 这两个是可以肯定的，赋初值
        rights[n] = n;

        // 求a[i]左边 连续超过a[i]高度的 最远的 下标，存到 lefts 中
        for (int i = 1; i <= n; i++)
        {
            t = i;
            while (t > 1 && a[i] <= a[t-1])  t = lefts[t - 1];      // 更新到 第t - 1 个数的下标
            lefts[i] = t;
        }

        // 求a[i]右边 连续超过a[i]高度的 最远的 下标，存到 rights 中
        for (int i = n - 1; i >= 1; i--)
        {
            t = i;
            while (t < n && a[i] <= a[t + 1])   t = rights[t + 1];      // 更新到 第t + 1 个数的下标
            rights[i] = t;
        }
        max_area = -9999;
        for (int i = 1; i <= n; i++)
        {
            max_area = max((rights[i] - lefts[i] + 1) * a[i] , max_area);
        }
        cout << max_area << endl;
    }
    return 0;
}