乘胜追击 树形DP

题意

分析

用$f[u][3]$表示三种状态：单独攻下 全部兵力攻下 协助攻下
然后进行状态转移即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 50010,M = N * 2;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
	char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
	while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int h[N],ne[M],e[M],idx;
ll f[N][3]; //单独攻下 全部兵力攻下 协助攻下
ll a[N],d[N];
int n;

void add(int x,int y){
	ne[idx] = h[x],e[idx] = y,h[x] = idx++;
}

void dfs(int u){
	f[u][0] = a[u] - d[u],f[u][1] = a[u];
	ll sum = 0,res = INF;
	for(int i = h[u];~i;i = ne[i]){
		int j = e[i];
		dfs(j);
		f[u][0] += min(f[j][0],min(f[j][1],f[j][2]));
		f[u][1] += min(f[j][0],min(f[j][1],f[j][2]));
		sum += min(f[j][0],min(f[j][1],f[j][2]));
	}
	for(int i = h[u];~i;i = ne[i]){
		int j = e[i];
		res = min(sum - min(f[j][0],min(f[j][1],f[j][2])) + f[j][1],res);
	}
	f[u][2] = res;
}


int main() {
	int T;
	read(T);
	while(T--){
		read(n);
		idx = 0;
		for(int i = 1;i <= n;i++){
			h[i] = -1;
			f[i][0] = f[i][1] = f[i][2] = INF;
		}
		for(int i = 2;i <= n;i++){
			int x;
			read(x);
			add(x,i);
		}
		for(int i = 1;i <= n;i++) read(a[i]);
		for(int i = 1;i <= n;i++) read(d[i]);
		dfs(1);
		dl(min(f[1][0],min(f[1][1],f[1][2])));
	}
	return 0;
}