题目描述

输入格式

输出格式

样例数据

2
3
1 1
1 2 3
2
1
1 2

28
9

max( S[u,p] , S[v,p] ) = max( S[1,u] , S[1,v] ) - ( S[1,p] - a[p] )

显然可以在O(n)内求出S[1,k]

S[1,i]对答案的贡献=

S[1,p]-a[p] 对答案的贡献为 = - ( p 作为 lca 的次数 * (S[1,p]-a[p]) )

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
//#include<cti1me>
#include<string.h>
#include<math.h>
#include<list>using namespace std;#define ll long long
#define pii pair<int,int>const int inf = 1e9+7;const int N = 100000 + 5;struct Edge{int to,next;
}edge[N];
int head[N];int a[N];void addEdge(int k,int u,int v){edge[k].to=v;edge[k].next=head[u];head[u]=k;
}ll s[N];
ll childs[N];
ll tS[N];void dfs(int u,int sum){s[u]=sum+a[u];childs[u]=1;for(int i=head[u];i!=-1;i=edge[i].next){int v=edge[i].to;dfs(v,s[u]);childs[u]+=childs[v];}
}ll f(int u){//u作为lca有多少种情况ll ans=1;for(int i=head[u];i!=-1;i=edge[i].next){int x=edge[i].to;ans+=childs[x];for(int j=edge[i].next;j!=-1;j=edge[j].next){int y=edge[j].to;ans+=childs[x]*childs[y];}}return ans;
}ll slove(int n){dfs(1,0);ll ans=0;for(int i=1;i<=n;++i){tS[i]=s[i];}sort(s+1,s+n+1);for(int i=1;i<=n;++i){ans+=s[i]*i;ans-=f(i)*(tS[i]-a[i]);}return ans;
}int main()
{//freopen("/home/lu/Documents/r.txt","r",stdin);//freopen("/home/lu/Documents/w.txt","w",stdout);int T;scanf("%d",&T);while(T--){int n;scanf("%d",&n);fill(head,head+n+1,-1);int nume=0;for(int i=2;i<=n;++i){int p;scanf("%d",&p);addEdge(nume++,p,i);}ll sub=0;for(int i=1;i<=n;++i){scanf("%d",&a[i]);sub+=a[i];}printf("%lld\n",2*slove(n)-sub);}return 0;
}