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 <bits/stdc++.h>
#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 <typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template <typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template <typename A, typename B, typename C>
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<int> 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<int>::iterator it = s[i].begin();
*it > 0 ? maximum += *it : minimum += *it;
}
else if(s[i].size() >= 2){
set<int>::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
", maximum, minimum);
}
return 0;
}