[十二省联考2019]春节十二响

https://www.luogu.org/problemnew/show/P5290

神tm一道蓝题，我考试的时候都在想啥...

考虑一条链，显然你是把两个链分别的最大值放在一起，次大值放在一起，等等

那么如果有多个链呢？你就把第一个链和第二个链按上面的操作，得到的新的结果再和第三个链合并...

正确性挺显然的...qwq

复杂度O(nlog^2n)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
#define O(x) cout << #x << " " << x << endl;
#define B cout << "breakpoint" << endl;
inline int read()
{
    int ans = 0,op = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-') op = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        (ans *= 10) += ch - '0';
        ch  = getchar();
    }
    return ans * op;
}
typedef long long ll;
const int maxn = 2e5 + 5;
struct node
{
    int to,next,cost;
}e[maxn << 1];
int fir[maxn],alloc;
void adde(int u,int v)
{
    e[++alloc].next = fir[u];
    fir[u] = alloc;
    e[alloc].to = v;
    swap(u,v);
    e[++alloc].next = fir[u];
    fir[u] = alloc;
    e[alloc].to = v;
}
int a[maxn];
int id[maxn],cnt,tp[maxn];
priority_queue<int> q[maxn];
void dfs(int u,int fa)
{
    //O(u);
    id[u] = ++cnt;
    for(int i = fir[u];i;i = e[i].next)
    {
        int v = e[i].to;
        if(v == fa) continue;
        dfs(v,u);
        if(q[id[u]].size() < q[id[v]].size()) swap(id[u],id[v]);
        int tot = q[id[v]].size();
        for(int i = 1;i <= tot;i++)
        {
            tp[i] = max(q[id[u]].top(),q[id[v]].top());
            q[id[u]].pop(),q[id[v]].pop();
        }
        for(int i = 1;i <= tot;i++) q[id[u]].push(tp[i]);
    }
    q[id[u]].push(a[u]);
}
int main()
{
    int n = read();
    for(int i = 1;i <= n;i++) a[i] = read();
    for(int i = 2;i <= n;i++) {int f = read(); adde(f,i);}
    dfs(1,0);
    ll ans = 0;
    while(q[id[1]].size()) ans += q[id[1]].top(),q[id[1]].pop();
    printf("%lld",ans);
}