Split The Tree（dfs序+树状数组）

Split The Tree

时间限制: 1 Sec  内存限制: 128 MB
提交: 46  解决: 11
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题目描述

You are given a tree with n vertices, numbered from 1 to n. ith vertex has a value wi
We define the weight of a tree as the number of different vertex value in the tree.
If we delete one edge in the tree, the tree will split into two trees. The score is the sum of these two trees’ weights.
We want the know the maximal score we can get if we delete the edge optimally.

输入

Input is given from Standard Input in the following format:
n
p2 p3  . . . pn
w1 w2  . . . wn
Constraints
2 ≤ n ≤ 100000 ,1 ≤ pi < i
1 ≤ wi ≤ 100000(1 ≤ i ≤ n), and they are integers
pi means there is a edge between pi and i

输出

Print one number denotes the maximal score.

样例输入

3
1 1
1 2 2

样例输出

3

 思路：

把序列延长一倍，每段子树的两边的区间连起来，将问题转化成维护区间中不同数字个数的最大值。

#include <bits/stdc++.h>
using namespace std;
const int maxn=2e5+10;
vector<pair<int,int> >vi[maxn];
int tree[maxn];
inline int lowbit(int x){return x&-x;}
void update(int x,int val) {for (int i=x; i<maxn; i+=lowbit(i)) tree[i]+=val;}
int Query(int x){int res=0;for (int i=x; i; i-=lowbit(i)) res+=tree[i]; return res;}
int st[maxn],dfn[maxn],ed[maxn],tot=0;
struct Edge{int u,v,next;}e[maxn];
int head[maxn],cnt=0;
inline void add(int u,int v){e[cnt]=Edge{u,v,head[u]};head[u]=cnt++;}
int n,vis[maxn],w[maxn];
void dfs(int x,int fa=0){
    st[x]=++tot;
    dfn[tot]=x;
    dfn[n+tot]=x;
    for (int v,i=head[x]; ~i; i=e[i].next){
        v=e[i].v;
        if(v==fa) continue;
        dfs(v,x);
    }
    ed[x]=tot;
}
int ans[maxn],res=0;
int main(){
    memset(head,-1,sizeof(head));
    scanf("%d",&n);
    for (int p,i=2; i<=n; i++) scanf("%d",&p),add(p,i),add(i,p);
    for (int i=1; i<=n; i++) scanf("%d",&w[i]);
    dfs(1);
    vi[ed[1]].push_back({st[1],1});
    for (int i=2; i<=n; i++){
        vi[ed[i]].push_back({st[i],i});
        vi[st[i]-1+n].push_back({ed[i]+1,i});
    }
    for (int i=1; i<=n+n; i++){
        if(vis[w[dfn[i]]]) update(vis[w[dfn[i]]],-1);
        update(i,1);
        vis[w[dfn[i]]]=i;
        for (auto u:vi[i])
            ans[u.second]+=Query(i)-Query(u.first-1);
    }
    for (int i=1; i<=n; i++) res=max(res,ans[i]);
    printf("%d\n",res);
    return 0;
}

View Code