P3047 [USACO12FEB]Nearby Cows G（树形dp）

[USACO12FEB]Nearby Cows G - 洛谷https://www.luogu.com.cn/problem/P3047一个很有趣的树形dp

我们可以设状态f【i】【j】， i为当前根，j为距离为j时的点权和

首先我们可以取1为根跑一遍dfs，将以i为根的子树的点权和记录下即用儿子更新父亲，此时1肯定已经求完了，我们就可以再一次从1开始跑dfs用父亲去更新儿子。

#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;
}