Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P6246 [IOI2000] 邮局 加强版

P6246 [IOI2000] 邮局 加强版

可以发现关于邮局的数量是凸函数。套一层 wqs 后,2D 决策单调性变成了 1D 决策单调性。使用队列维护三元组解决。同样的,还需要维护 $cnt$ ,同样的函数值,取 $cnt$ 小的。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 5e5 + 10;
struct Node
{
int p, l, r;
} q[N];
int hd, tl;
struct Data
{
int t;
LL s;
friend bool operator<(const Data &x, const Data &y)
{
return x.s < y.s || (x.s == y.s && x.t < y.t);
}
friend Data operator+(const Data &x, const LL &k)
{
return (Data){x.t + 1, x.s + k};
}
} f[N];
LL s[N];
int n, m, d[N];
Data cal(int k, int i)
{
int mid = i + k + 1 >> 1;
return f[k] + (s[i] + s[k] - 2 * s[mid] + (i + k & 1) * d[mid]);
}
void check(LL x)
{
hd = 1, tl = 0;
q[++tl] = (Node){0, 1, n};
for (int i = 1; i <= n; i++)
{
while (hd <= tl && q[hd].r < i)
hd++;
f[i] = cal(q[hd].p, i);
f[i].s += x;
int p = n + 1;
while (hd <= tl && cal(i, q[tl].l) < cal(q[tl].p, q[tl].l))
p = q[tl--].l;
if (hd <= tl && cal(i, q[tl].r) < cal(q[tl].p, q[tl].r))
{
int l = q[tl].l, r = q[tl].r;
while (l <= r)
{
int mid = l + r >> 1;
cal(i, mid) < cal(q[tl].p, mid) ? (p = mid, r = mid - 1) : l = mid + 1;
}
q[tl].r = p - 1;
}
if (p <= n)
q[++tl] = (Node){i, p, n};
}
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%d", &d[i]);
sort(d + 1, d + n + 1);
for (int i = 1; i <= n; i++)
s[i] = s[i - 1] + d[i];
LL l = 0, r = 5e11, res;
while (l <= r)
{
LL mid = l + r >> 1;
check(mid);
f[n].t <= m ? (res = f[n].s - mid * m, r = mid - 1) : l = mid + 1;
}
printf("%lld", res);
return 0;
}