贪心

应该不是正解。。用set维护每个节点孩子的权值，然后再维护一个次大值和次小值。

这里的次值非常好维护，对于次大值而言，小于0那就是0，不然就是每个set的倒数第二个元素，因为最大值如果大于0是肯定要选的，最小值同理。。

然后这是用scanf卡过去的。。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();    return w ? -ret : ret;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template 
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template 
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template 
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 100005;int _, n, w[N], p[N];multiset
        
          s[N];int main(){    for(scanf("%d", &_); _; _ --){        scanf("%d", &n);        for(int i = 0; i <= n; i ++) s[i].clear(), p[i] = 0;        for(int i = 2; i <= n; i ++) scanf("%d", &p[i]);        for(int i = 1; i <= n; i ++) scanf("%d", &w[i]);        for(int i = 1; i <= n; i ++){            s[p[i]].insert(w[i]);        }        ll maximum = 0, minimum = 0, x = 0, y = 0;        for(int i = 0; i <= n; i ++){            if(s[i].size() == 1){                set
         
          ::iterator it = s[i].begin(); *it > 0 ? maximum += *it : minimum += *it; } else if(s[i].size() >= 2){ set
          
           ::iterator it = --s[i].end(); if(*it > 0) maximum += *it; if(*(--it) > 0) x = max(x, (*it) * 1LL); it = s[i].begin(); if(*it < 0) minimum += *it; if(*(++it) < 0) y = min(y, (*it) * 1LL); } } maximum += x, minimum += y; printf("%lld %lld\n", maximum, minimum); } return 0;}