python/TangDou/Topic/HuanGenDp/P3047.cpp

#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];
struct ljj {
    ll to, stb;
} a[200010];
ll s = 0;
inline void insert(ll x, ll 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;
        if (xx == fa)
            continue;
        dfs(xx, x);
        for (re ll 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;
        // 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--)
            dp[xx][j] -= dp[xx][j - 2]; // 简单容斥
        for (re ll 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();
        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；
    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
        printf("%lld\n", ans);
    }
    return 0;
}