标签：acm   hdu   动态规划   dp   

Largest Rectangle in a Histogram

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.

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

8
4000

题意  求条形图中最大矩形的面积  输入给你条的个数  每个条的高度hi   （以下等于也视为高）

只要知道第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);

但是直接枚举每一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\n", ans);
    }
    return 0;
}

可以发现   当第i-1个比第i个高的时候   比第i-1个高的所有也一定比第i个高  

于是可以用到动态规划的思想

令left[i]表示包括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]  从前往后可以递推出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\n", ans);
    }
    return 0;
}

HDU 1506 Largest Rectangle in a Histogram(DP),布布扣,bubuko.com

HDU 1506 Largest Rectangle in a Histogram(DP)

标签：acm   hdu   动态规划   dp   

原文地址：http://blog.csdn.net/iooden/article/details/38379065