hdu1506(Largest Rectangle in a Histogram)单调队列实现 (广告印刷变形)_�少_新浪博客


首先考虑问题需要的时间复杂度如果是on)的算法,单调队列是首选。
最终的广告牌一定等于某个建筑物的高度×其能达到的最大长度
现在,建筑物的高度已知,现在只需要知道每个高度能达到的最大长度是多少。由于n100000,我们只能用On)或Onlogn)的算法。当让完全可以使用rmq(本人太弱还没研究ST算法所以木用 , 稍后一定补上为大家奉上),在后边的论文中会讲到。
现在讲时间复杂度为on)的单调队列的方法。
继续上边的思路,对于每个建筑物,只需要找到其能够扩展到的最大宽度即可。也就是这个建筑物的左右两边的比它低或等于它的建筑物个数。
如何用单调队列呢?
1~n依次进队,维护一个单调递增的序列(最小队列) (注意que要存 高度方便比较 和 方块的起始位置 , 方便更新 r )。
每次加入元素后按照单调队列的入队性质维护其单调性,当然这样做必然会使一些元素出队,出队的元素一定要比当前加入的元素大也就是说当前元素就是出队的元素能在右侧达到的最远的建筑物! 然后记录当前元素i-1的位置(距离) 。 

注意,要让
l[1~n]一开始要初始化为n r[1~n]初始化为1 处理极限(就是如果没出过队就是覆盖所有长度) 并且让该元素入队一次(会使当前队列中的所有元素出队),保证每个元素都有其“右极限”的值.
要求“左极限”同理,只需从n~1循环即可
这道题是对单调队列的变形使用。由于问题的结果具有单调性很好的利用出队元素的特性。

long long n, h[100001], r[100001], l[100001], pos[100001], val[100001];
 7 int main()
 8 {
 9     int head, tail;
10     while(scanf("%ld", &n), n)
11     {
12         head = 0, tail = -1;
13         for(int i=1 ; i<=n ; i++)
14         {
15             scanf("%ld", &h[i]);
16             l[i] = 1;
17             r[i] = n;
18             while(head <= tail && h[i] < val[tail])
19             {
20                 r[pos[tail]] = i-1;
21                 --tail;
22             }
23             pos[++tail] = i;
24             val[tail] = h[i];
25         }
26         head = 0, tail = -1;
27         long long int res = 0;
28         for(int i=n ; i>=1 ; i--)
29         {
30             while(head <= tail && h[i] < val[tail])
31             {
32                 l[pos[tail]] = i+1;
33                 --tail;
34             }
35             pos[++tail] = i;
36             val[tail] = h[i];
37         }
38         for(int i=1 ; i<=n ; i++)
39             res = max(res, (r[i] - l[i] + 1)*h[i]);
40         cout << res << endl;
41     }
42 }


Also check http://www.cnblogs.com/baijin/archive/2013/06/16/3116138.html
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