CF1313C2 Skyscrapers (hard version)

思路：

使用单调栈。

实现：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int N = 500005;
 5 ll a[N], l[N], r[N];
 6 int main()
 7 {
 8     int n;
 9     while (cin >> n)
10     {
11         stack<int> st;
12         for (int i = 1; i <= n; i++) cin >> a[i];
13         for (int i = 1; i <= n; i++)
14         {
15             while (!st.empty() && a[st.top()] >= a[i]) st.pop();
16             if (st.empty()) l[i] = i * a[i];
17             else l[i] = l[st.top()] + (i - st.top()) * a[i];
18             st.push(i);
19         }
20         while (!st.empty()) st.pop();
21         for (int i = n; i >= 1; i--)
22         {
23             while (!st.empty() && a[st.top()] >= a[i]) st.pop();
24             if (st.empty()) r[i] = (n - i + 1) * a[i];
25             else r[i] = r[st.top()] + (st.top() - i) * a[i];
26             st.push(i);
27         }
28         ll res = 0, id = 0;
29         for (int i = 1; i <= n; i++)
30         {
31             ll tmp = l[i] + r[i] - a[i];
32             if (tmp > res) { res = tmp; id = i; }
33         }
34         for (int i = id - 1; i >= 1; i--) a[i] = min(a[i], a[i + 1]);
35         for (int i = id + 1; i <= n; i++) a[i] = min(a[i], a[i - 1]);
36         for (int i = 1; i <= n; i++) cout << a[i] << " ";
37         cout << endl;
38     }
39     return 0;
40 }

标签：CF1313C2,Skyscrapers,int,top,hard,pop,st,id,empty
来源： https://www.cnblogs.com/wangyiming/p/12355937.html