poj 2559 单调栈 | himdd


题意:给出一系列的1*h的矩形,求矩形的最大面积。
如图:
可转化为求找一个子序列。 使得这个序列的长度乘以序列最小数最大。
分析:这是一个单调栈的问题,维护栈单调不减。 单调栈 主要是大家要自己枚举,需要找到每个元素 最左能扩展到那 ,最右能扩展到那,当然最小的是你枚举的那个元素。有些细节问题我标有"注意"的地方自己好好想想。和poj 2796有相似的地方。主要是范围不一样,也就是栈的下标有变化了。
const int A=100005;
09struct{
10    long long h;
11    int id;
12}st[A];
13long long h[A];
14int main()
15{
16    //freopen("1.txt","r",stdin);
17    int n;
18    while(scanf("%d",&n),n)
19    {
20        for(int i=1;i<=n;i++)
21        {
22            scanf("%d",&h[i]);
23        }
24        h[n+1]=0;
25        long long ans=0,tmp;
26        int top=1;
27        st[top].h=h[1];
28        st[top].id=1;
29        for(int i=2;i<=n+1;i++)
30        {
31            if(st[top].h>h[i])
32            {
33                while(top!=0&&st[top].h>h[i])
34                {
35                    tmp=st[top].h*(i-1-st[top].id+1);
36                    //st[top].id~i-1的h都是大于等于st[top].h
37                    //栈顶元素是序列的最小元素
38                    if(ans<tmp)
39                    {
40                        ans=tmp;
41                    }
42                    top--;
43                }
44                top++;
45                st[top].h=h[i];
46                //注意这里st[top].id还是原来的id,因为st[top].id~i-1的h都是>h[i]
47                //下一次以h[i]为高计算时,st[top].id~i-1也是要加上的。
48            }
49            else//满足栈单调不减入栈
50            {
51                top++;
52                st[top].h=h[i];
53                st[top].id=i;//注意这里的下标还是本来的i
54            }
55        }
56        printf("%lld\n",ans);
57    }
58}
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