icpc2020南京站 M.Monster Hunter (树形背包)

题目链接：https://ac.nowcoder.com/acm/problem/216013

设 \(dp[i][j][0/1]\) 表示杀到第 \(i\) 个节点，已经用魔法杀了 \(j\) 个怪物，当前节点使不使用魔法，所消耗的最小能量

注意下树形背包的写法

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
typedef long long ll;

const int maxn = 2010; 
const ll inf = 1ll << 60ll;

int T, n, m;
int a[maxn];
ll tmp[maxn][2], dp[maxn][maxn][2];

int h[maxn], cnt = 0;
struct E{
	int to, next;
}e[maxn << 1];
void add(int v, int u){
	e[++cnt].next = h[u];
	e[cnt].to = v;
	h[u] = cnt;
}

int sz[maxn];

void dfs(int u, int par){
	sz[u] = 1;
	dp[u][0][0] = a[u];
	dp[u][1][1] = 0;
	for(int i = h[u] ; i != -1 ; i = e[i].next){
		int v = e[i].to;
		if(v == par) continue;
		
		dfs(v, u);
		
		for(int j = 0 ; j <= sz[u] + sz[v] ; ++j) tmp[j][0] = tmp[j][1] = inf;
		for(int j = 0 ; j <= sz[u] ; ++j){
			for(int k = 0 ; k <= sz[v] ; ++k){
				tmp[j + k][0] = min(tmp[j + k][0], dp[u][j][0] + dp[v][k][0] + a[v]);
                tmp[j + k][0] = min(tmp[j + k][0], dp[u][j][0] + dp[v][k][1]);
                tmp[j + k][1] = min(tmp[j + k][1], dp[u][j][1] + dp[v][k][0]);
                tmp[j + k][1] = min(tmp[j + k][1], dp[u][j][1] + dp[v][k][1]);
			}
		}
		
		for(int j = 0 ; j <= sz[u] + sz[v] ; ++j){
			dp[u][j][1] = tmp[j][1];
			dp[u][j][0] = tmp[j][0];
		}
		
		sz[u] += sz[v];
	}
}

ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < '0' || ch > '9'){ if(ch == '-') f = -1; ch = getchar(); } while(ch >= '0' && ch <= '9'){ s = s * 10 + ch - '0'; ch = getchar(); } return s * f; }

int main(){
	T = read();
	
	while(T--){
		memset(h, -1, sizeof(h));
		n = read();	
		
		for(int i = 0 ; i <= n ; ++i){
			for(int j = 0 ; j <= n ; ++j){
				dp[i][j][0] = dp[i][j][1] = inf;
			} 
		} 

		cnt = 0;
		
		int x;
		for(int i = 2 ; i <= n ; ++i){
			x = read();
			add(i, x), add(x, i);
		}
		for(int i = 1 ; i <= n ; ++i) a[i] = read();
		
		dfs(1, 0);
		
		for(int i = 0 ; i <= n ; ++i) printf("%lld ", min(dp[1][i][0], dp[1][i][1]));
		printf("\n");
	}

	return 0;
}