Blog of RuSun

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

P3047 [USACO12FEB]Nearby Cows G

P3047 [USACO12FEB]Nearby Cows G

$f _ {i, j}$ 表示 $i$ 点中深度为 $j$ 的点权和。

第一次 $dfs$ ,以 $1$ 为根,答案很好求。第二次 $dfs$ 时,我们只保证当前的 $x$ 的 $f$ 有意义。我们以一个新的点为根,原来的根的答案要减去该点的贡献,同时该点加上原来的根的贡献。回溯时撤销更改。

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
#include <cstdio>
using namespace std;
const int N = 1e5 + 10, M = N << 1, K = 30;
int n, k, c[N], f[N][K], s[N];
int idx, hd[N], nxt[M], edg[M];
void dfs1(int x, int fa)
{
f[x][0] = c[x];
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != fa)
{
dfs1(edg[i], x);
for (int j = 1; j <= k; j++)
f[x][j] += f[edg[i]][j - 1];
}
}
void dfs2(int x, int fa)
{
for (int i = 0; i <= k; i++)
s[x] += f[x][i];
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != fa)
{
for (int j = 1; j <= k; j++)
f[x][j] -= f[edg[i]][j - 1];
for (int j = 1; j <= k; j++)
f[edg[i]][j] += f[x][j - 1];
dfs2(edg[i], x);
for (int j = 1; j <= k; j++)
f[edg[i]][j] -= f[x][j - 1];
for (int j = 1; j <= k; j++)
f[x][j] += f[edg[i]][j - 1];
}
}
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
int main()
{
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1, u, v; i < n; i++)
{
scanf("%d%d", &u, &v);
add(u, v);
add(v, u);
}
for (int i = 1; i <= n; i++)
scanf("%d", &c[i]);
dfs1(1, 0);
dfs2(1, 0);
for (int i = 1; i <= n; i++)
printf("%d\n", s[i]);
return 0;
}