小A的柱状图

思路

经典的单调栈题目，对于每一个位置，我们维护他以当前高度可以到达的最左方，以及他当前高度可以到达的最有方，显然就有以他的高度的矩形块的面积就出来了，所以我们只需要统计n个矩形的最大值就行。

具体细节操作看代码注释。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 1e6 + 10;int h[N], len[N], l[N], r[N], n;stack<int> stk;int main() {//freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);n = read();for(int i = 1; i <= n; i++) {len[i] = len[i - 1] + read();}for(int i = 1; i <= n; i++) {h[i] = read();}for(int i = 1; i <= n; i++) {while(stk.size() && h[stk.top()] >= h[i]) stk.pop();if(stk.empty()) l[i] = 1;else l[i] = stk.top() + 1;stk.push(i);}while(stk.size()) stk.pop();for(int i = n; i >= 1; i--) {while(stk.size() && h[stk.top()] >= h[i]) stk.pop();if(stk.empty()) r[i] = n;else r[i] = stk.top() - 1;stk.push(i);}ll ans = 0;for(int i = 1; i <= n; i++) {ans = max(ans, 1ll * (len[r[i]] - len[l[i] - 1]) * h[i]);}printf("%lld\n", ans);return 0;
}