题目链接:nowcoder多校4-sequence
题意:
给定两个数组a和b,然后询问一个区间最大值,区间最大值的定义为该区间内a数组中的最小值与b数组中区间和的乘积。
题解:
单调栈+线段树。
对于a[i],求出a[i]之前第一个比a[i]小的a[l],a[i]之后第一个比a[i]小的a[r],所以在区间[l+1,r-1]中的最小值即为a[i]。
然后对于数组b,记录其前缀和,然后对于a[i],求出区间[i,r-1]中的最大值与区间[l,i-1]中的最小值,作差后与a[i]相乘即可。区间最大值最小值查询可使用线段树/RMQ等。
参考代码:
#include <bits/stdc++.h>
#define IL inline
using namespace std;
typedef long long ll;
typedef double db;
const int MAXN = 3e6 + 6;
const ll inf = 0x3f3f3f3f;
int n;
int a[MAXN], b[MAXN], lm[MAXN], rm[MAXN];
ll sum[MAXN];
ll maxv[MAXN << 2], minv[MAXN << 2];
void pushup(int rt)
{
minv[rt] = min(minv[rt << 1], minv[rt << 1 | 1]);
maxv[rt] = max(maxv[rt << 1], maxv[rt << 1 | 1]);
}
void build(int l, int r, int rt)
{
if (l == r)
{
minv[rt] = maxv[rt] = sum[l];
return;
}
int mid = (l + r) / 2;
build(l, mid, rt << 1);
build(mid + 1, r, rt << 1 | 1);
pushup(rt);
}
ll query_max(int L, int R, int l, int r, int rt)
{
if (L <= l && r <= R)
return maxv[rt];
ll res = -inf;
int mid = (l + r) / 2;
if (L <= mid)
res = max(res, query_max(L, R, l, mid, rt << 1));
if (R > mid)
res = max(res, query_max(L, R, mid + 1, r, rt << 1 | 1));
return res;
}
ll query_min(int L, int R, int l, int r, int rt)
{
if (L <= l && r <= R)
return minv[rt];
ll res = inf;
int mid = (l + r) / 2;
if (L <= mid)
res = min(res, query_min(L, R, l, mid, rt << 1));
if (R > mid)
res = min(res, query_min(L, R, mid + 1, r, rt << 1 | 1));
return res;
}
struct node
{
int num, id;
};
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &a[i]);
sum[0] = 0;
for (int i = 1; i <= n; i++)
scanf("%d", &b[i]), sum[i] = sum[i - 1] + b[i];
build(0, n, 1);
stack<node> sta;
a[0] = a[n + 1] = -inf;
for (int i = 0; i <= n; i++)
{
node now;
now.num = a[i], now.id = i;
if (sta.empty())
{
sta.push(now);
lm[i] = i - 1;
continue;
}
while (!sta.empty())
{
if (a[i] <= sta.top().num)
sta.pop();
else
break;
}
if (sta.empty())
{
sta.push(now);
lm[i] = i - 1;
}
else
{
lm[i] = sta.top().id;
sta.push(now);
}
}
while (!sta.empty())
sta.pop();
for (int i = n + 1; i >= 1; i--)
{
node now;
now.num = a[i], now.id = i;
if (sta.empty())
{
sta.push(now);
rm[i] = i + 1;
continue;
}
while (!sta.empty())
{
if (a[i] <= sta.top().num)
sta.pop();
else
break;
}
if (sta.empty())
{
sta.push(now);
rm[i] = i + 1;
}
else
{
rm[i] = sta.top().id;
sta.push(now);
}
}
ll ans = -inf;
for (int i = 1; i <= n; i++)
{
ll temp, minn, maxn;
if (a[i] >= 0)
{
maxn = query_max(i, rm[i] - 1, 0, n, 1);
minn = query_min(lm[i], i - 1, 0, n, 1);
temp = maxn - minn;
}
else
{
maxn = query_max(lm[i], i - 1, 0, n, 1);
minn = query_min(i, rm[i] - 1, 0, n, 1);
temp = minn - maxn;
}
ans = max(ans, temp * (ll)a[i]);
}
printf("%lld\n", ans);
return 0;
}