HDU 1506 Largest Rectangle in a Histogram(DP)

Largest Rectangle in a Histogram

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11137 Accepted Submission(s): 3047

Problem 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, ..., 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

题意求条形图中最大矩形的面积输入给你条的个数每一个条的高度hi （下面等于也视为高）

仅仅要知道第i个条左边连续多少个（a）比他高右边连续多少个（b）比他高那么以这个条为最大高度的面积就是hi*(a+b+1);

可是直接枚举每个的话肯定会超时的超时代码

#include<cstdio>
using namespace std;
const int N = 100005;
typedef long long ll;
ll h[N]; int n,wide[N];
int main()
{
    while (scanf ("%d", &n), n)
    {
        for (int i = 1; i <= n; ++i)
            scanf ("%I64d", &h[i]);
        for (int i = 1; i <= n; ++i)
        {
            wide[i] = 1;
            int k = i;
            while (k > 1 && h[--k] >= h[i]) ++wide[i];
            k = i;
            while (k < n && h[++k] >= h[i]) ++wide[i];
        }
        ll ans = 0;
        for (int i = 1; i <= n; ++i)
            if (h[i]*wide[i] > ans) ans = h[i] * wide[i];
        printf ("%I64d
", ans);
    }
    return 0;
}

能够发现当第i-1个比第i个高的时候比第i-1个高的全部也一定比第i个高

于是能够用到动态规划的思想

令left[i]表示包含i在内比i高的连续序列中最左边一个的编号 right[i]为最右边一个的编号

那么有当h[left[i]-1]>=h[i]]时 left[i]=left[left[i]-1] 从前往后能够递推出left[i]

同理当h[right[i]+1]>=h[i]]时 right[i]=right[right[i]+1] 从后往前可递推出righ[i]

最后答案就等于 max((right[i]-left[i]+1)*h[i])了

#include<cstdio>
using namespace std;
const int N = 100005;
typedef long long ll;
ll h[N];
int n, left[N], right[N];
int main()
{
    while (scanf ("%d", &n), n)
    {
        for (int i = 1; i <= n; ++i)
            scanf ("%I64d", &h[i]), left[i] = right[i] = i;
        h[0] = h[n + 1] = -1;
        for (int i = 1; i <= n; ++i)
            while (h[left[i] - 1] >= h[i])
                left[i] = left[left[i] - 1];
        for (int i = n; i >= 1; --i)
            while (h[right[i] + 1] >= h[i])
                right[i] = right[right[i] + 1];
        ll ans = 0;
        for (int i = 1; i <= n; ++i)
            if (h[i] * (right[i] - left[i] + 1) > ans) ans = h[i] * ll (right[i] - left[i] + 1);
        printf ("%I64d
", ans);
    }
    return 0;
}