poj2559

题目链接：http://poj.org/problem?id=2559
解法：单调栈
这个题是 https://blog.youkuaiyun.com/qq_41646772/article/details/96701780 的简化版，维护一个单调递增的栈，栈内存下标。这题没什么好说的，看代码就能明白

#include <cstdio>
#include <algorithm>
#include <stack>
using namespace std;
typedef long long LL;
int N;
LL h[100005],ans;
stack<int>sta;
int main()
{
    while(~scanf("%d",&N)&&N)
    {
        while(!sta.empty())sta.pop();
        for(int i=0;i<N;i++)scanf("%lld",&h[i]);
        h[N]=-1;
        ans=0;
        for(int i=0;i<=N;i++)
        {
            if(sta.empty()||h[i]>=h[sta.top()])sta.push(i);
            else
            {
                int top;
                while(!sta.empty()&&h[i]<h[sta.top()])
                {
                    top=sta.top();
                    sta.pop();
                    LL x=i-top,y=h[top];
                    ans=max(ans,x*y);
                }
                sta.push(top);
                h[top]=h[i];
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}