Blog of RuSun

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

P5904 [POI2014]HOT-Hotels 加强版

P5904 [POI2014]HOT-Hotels 加强版

$f _ {u, k}$ 表示在 $u$ 子树中,与 $u$ 距离为 $k$ 的点的个数,$g _ {u, k}$ 表示点对 $(x, y)$ 满足 $dis(lca(x, y), x) = dis(lca(x, y), y) = dis(lca(x, y), u) + k$ 的数量(也就是还需要补足 $k$ 距离),可以长链剖分优化。

$g$ 是倒序传递的,因此需要前后各需要开一倍空间。

查看代码
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
#include <cstdio>
#include <vector>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 1e5 + 10;
vector <int> g[N];
int n, d[N], son[N];
LL ans, v[N << 2], *f[N], *h[N], *cur = v;
void dfs1 (int x, int fa)
{
for (int i : g[x])
if (i ^ fa)
{
dfs1(i, x);
(d[i] > d[son[x]]) && (son[x] = i);
}
d[x] = d[son[x]] + 1;
}
void dfs2 (int x, int fa)
{
if (son[x])
{
f[son[x]] = f[x] + 1, h[son[x]] = h[x] - 1;
dfs2(son[x], x);
}
f[x][0] = 1, ans += h[x][0];
for (int i : g[x])
if (i ^ fa && i ^ son[x])
{
f[i] = cur, cur += d[i] << 1;
h[i] = cur, cur += d[i];
dfs2(i, x);
for (int k = 0; k < d[i]; k++)
{
k && (ans += f[x][k - 1] * h[i][k]);
ans += h[x][k + 1] * f[i][k];
}
for (int k = 0; k < d[i]; k++)
{
h[x][k + 1] += f[x][k + 1] * f[i][k];
k && (h[x][k - 1] += h[i][k]);
f[x][k + 1] += f[i][k];
}
}
}
int main ()
{
read(n);
for (int i = 1, a, b; i < n; i++)
{
read(a), read(b);
g[a].push_back(b);
g[b].push_back(a);
}
dfs1(1, 0);
f[1] = cur, cur += d[1] << 1;
h[1] = cur, cur += d[1];
dfs2(1, 0);
write(ans);
return 0;
}