单调栈

http://poj.org/problem?id=2559
 1 a[n + 1] = p = 0;
 2 for (int i = 1; i <= n + 1; i++) 
 3 {
 4     if (a[i] > s[p]) 
 5     {
 6         s[++p] = a[i], w[p] = 1;
 7     } 
 8     else 
 9     {
10         int width=0;
11         while (s[p] > a[i]) 
12         {
13             width += w[p];
14             ans = max(ans, (long long)width * s[p]);
15             p--;
16         }
17         s[++p] = a[i], w[p] = width + 1;
18     }
19 }

 及时排除不可能的选项，保持决策集合的高度和秩序性

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原文地址：https://www.cnblogs.com/universeplayer/p/10655729.html