单调栈 + 线段树
先单调栈预处理a中每一个数覆盖的范围,然后用线段树维护b的前缀和,在选i的范围内查询区间值。
讨论一下a的正负性
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false)
using namespace std;
typedef long long LL;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
int ret = 0, w = 0; char ch = 0;
while(!isdigit(ch)){
w |= ch == '-', ch = getchar();
}
while(isdigit(ch)){
ret = (ret << 3) + (ret << 1) + (ch ^ 48);
ch = getchar();
}
return w ? -ret : ret;
}
inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
A ans = 1;
for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
return ans;
}
const int N = 4000005;
int n, a[N], b[N], L[N], R[N];
LL mx[N<<2], mn[N<<2], sum[N];
void calc(){
stack<int> st;
a[0] = a[n + 1] = -INF;
st.push(0);
for(int i = 1; i <= n; i ++){
while(!st.empty() && a[st.top()] >= a[i]) st.pop();
L[i] = st.top();
st.push(i);
}
while(!st.empty()) st.pop();
st.push(n + 1);
for(int i = n; i >= 1; i --){
while(!st.empty() && a[st.top()] >= a[i]) st.pop();
R[i] = st.top();
st.push(i);
}
}
void push_up(int rt){
mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);
}
void buildTree(int rt, int l, int r){
if(l == r){
mx[rt] = mn[rt] = sum[l];
return;
}
int mid = (l + r) >> 1;
buildTree(rt << 1, l, mid);
buildTree(rt << 1 | 1, mid + 1, r);
push_up(rt);
}
LL queryMax(int rt, int l, int r, int ql, int qr){
if(l == ql && r == qr){
return mx[rt];
}
int mid = (l + r) >> 1;
if(qr <= mid) return queryMax(rt << 1, l, mid, ql, qr);
else if(ql > mid) return queryMax(rt << 1 | 1, mid + 1, r, ql, qr);
else return max(queryMax(rt << 1, l, mid, ql, mid), queryMax(rt << 1 | 1, mid + 1, r, mid + 1, qr));
}
LL queryMin(int rt, int l, int r, int ql, int qr){
if(l == ql && r == qr){
return mn[rt];
}
int mid = (l + r) >> 1;
if(qr <= mid) return queryMin(rt << 1, l, mid, ql, qr);
else if(ql > mid) return queryMin(rt << 1 | 1, mid + 1, r, ql, qr);
else return min(queryMin(rt << 1, l, mid, ql, mid), queryMin(rt << 1 | 1, mid + 1, r, mid + 1, qr));
}
int main(){
n = read();
for(int i = 1; i <= n; i ++) a[i] = read();
for(int i = 1; i <= n; i ++) b[i] = read();
for(int i = 1; i <= n; i ++) sum[i] = sum[i - 1] + b[i];
calc();
buildTree(1, 0, n);
LL ans = 0;
for(int i = 1; i <= n; i ++){
int pl = L[i], pr = i - 1;
int sl = i, sr = R[i] - 1;
if(a[i] < 0){
ans = max(ans, 1LL * a[i] * (queryMin(1, 0, n, sl, sr) - queryMax(1, 0, n, pl, pr)));
}
else if(a[i] > 0){
ans = max(ans, 1LL * a[i] * (queryMax(1, 0, n, sl, sr) - queryMin(1, 0, n, pl, pr)));
}
}
printf("%lld
", ans);
return 0;
}