P3047 [usaco12feb]nearby cows g (tree DP)

[USACO12FEB]Nearby Cows G - Luogu https://www.luogu.com.cn/problem/P3047 A very interesting tree dp

We can set the State f【i】【j】, i Is the current root ,j The distance is j Point weight sum of time

First we can take 1 Run for the root dfs, Will be with i The point weight and record of the subtree for the root, that is, update the father with the son , here 1 It must be over , We can start from 1 Start running dfs Use the father to renew the son .

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cstring>
#include <set>
#include <cmath>
#include <map>
#include <bitset>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int MN = 65005;
const int MAXN = 1e6 + 10;
const int INF = 0x3f3f3f3f;
#define IOS ios::sync_with_stdio(false)
#define lowbit(x) ((x)&(-x))
using P = pair<int, int>;

int n, k;
int head[MAXN];
int ver[MAXN];
int nxt[MAXN];
int cost[MAXN];
int cnt;
int dep[MAXN];
int dp[MAXN][21];
void add(int x, int y) {
	ver[++cnt] = y;
	nxt[cnt] = head[x];
	head[x] = cnt;
}

void dfs1(int p, int fa) {
	for (int i = head[p]; i; i = nxt[i]) {
		int v = ver[i];
		if (v == fa)
			continue;
//		dep[v] = dep[p] + 1;
		dfs1(v, p);
		for (int j = 1; j <= k; j++) {
			dp[p][j] += dp[v][j - 1];
		}
	}
}

void dfs2(int p, int fa) {
	for (int i = head[p]; i; i = nxt[i]) {
		int v = ver[i];
		if (v == fa)
			continue;
		for (int j = k; j >= 2; j--) {
			dp[v][j] += dp[p][j - 1] - dp[v][j - 2];
		}
		dp[v][1] += cost[p];
		dfs2(v, p);
	}
}

int main() {
	scanf("%d %d", &n, &k);
	int x, y;
	for (int i = 1; i <= n - 1; i++) {
		scanf("%d %d", &x, &y);
		add(x, y);
		add(y, x);
	}
	for (int i = 1; i <= n; i++) {
		scanf("%d", cost + i);
		dp[i][0] = cost[i];
	}
	dfs1(1, 0);
	dfs2(1, 0);
	for (int i = 1; i <= n; i++) {
		int ans = 0;
		for (int j = 0; j <= k; j++) {
			ans += dp[i][j];
		}
		printf("%d\n", ans);
	}
	return 0;
}