main
黄海 2 years ago
parent 6e78c320df
commit 24504d62e2

@ -1,75 +1,60 @@
#include <bits/stdc++.h>
#define re register
typedef long long ll;
using namespace std;
inline ll read() {
ll a = 0, f = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') f = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
a = a * 10 + c - '0';
c = getchar();
}
return a * f;
} // 好用的快读
ll n, q;
ll head[200010], dp[200010][21];
#define int long long
#define endl "\n"
int n, q;
int head[200010], dp[200010][21];
struct ljj {
ll to, stb;
int to, stb;
} a[200010];
ll s = 0;
inline void insert(ll x, ll y) {
int s = 0;
inline void insert(int x, int y) {
s++;
a[s].stb = head[x];
a[s].to = y;
head[x] = s;
}
inline void dfs(ll x, ll fa) {
for (re ll i = head[x]; i; i = a[i].stb) {
ll xx = a[i].to;
inline void dfs(int x, int fa) {
for (int i = head[x]; i; i = a[i].stb) {
int xx = a[i].to;
if (xx == fa)
continue;
dfs(xx, x);
for (re ll j = 1; j <= q; j++)
for (int j = 1; j <= q; j++)
dp[x][j] += dp[xx][j - 1]; // 第一遍dp
}
}
inline void dfs1(ll x, ll fa) {
for (re ll i = head[x]; i; i = a[i].stb) {
ll xx = a[i].to;
if (xx == fa)
continue;
inline void dfs1(int x, int fa) {
for (int i = head[x]; i; i = a[i].stb) {
int xx = a[i].to;
if (xx == fa) continue;
// 在第一次遍历时 dp[1][2] 包括了 dp[2][1] 2的子树权值
// 然鹅 ans在统计dp[2][3] 的时候也加上了 dp[2][1] 2的子树权值
// 第二次遍历 dp[2][3] 又加上了 dp[2][1];
// 所以需要简单容斥一下;
for (re ll j = q; j >= 2; j--)
for (int j = q; j >= 2; j--)
dp[xx][j] -= dp[xx][j - 2]; // 简单容斥
for (re ll j = 1; j <= q; j++)
for (int j = 1; j <= q; j++)
dp[xx][j] += dp[x][j - 1]; // 第二遍dp
dfs1(xx, x);
}
}
int main() {
n = read();
q = read();
for (re ll i = 1; i < n; i++) {
ll x = read(), y = read();
signed main() {
cin >> n >> q;
for (int i = 1; i < n; i++) {
int x, y;
cin >> x >> y;
insert(x, y);
insert(y, x);
}
for (re ll i = 1; i <= n; i++)
dp[i][0] = read(); // 每个节点往外0距离就是它本身的权值
for (int i = 1; i <= n; i++) cin >> dp[i][0]; // 每个节点往外0距离就是它本身的权值
dfs(1, 0);
dfs1(1, 0);
for (re ll i = 1; i <= n; i++) {
ll ans = 0;
for (re ll j = 0; j <= q; j++)
ans += dp[i][j]; // ans统计答案
for (int i = 1; i <= n; i++) {
int ans = 0;
for (int j = 0; j <= q; j++) ans += dp[i][j]; // ans统计答案
printf("%lld\n", ans);
}
return 0;
}
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