HUD Crazy Bobo 树上dfs

Crazy Bobo

Problem Description

Bobo has a tree,whose vertices are conveniently labeled by 1,2,...,n.Each node has a weight wi. All the weights are distrinct.
A set with m nodes v1,v2,...,vm is a Bobo Set if:
- The subgraph of his tree induced by this set is connected.
- After we sort these nodes in set by their weights in ascending order,we get u1,u2,...,um,(that is,wui<wui+1 for i from 1 to m-1).For any node x in the path from ui to ui+1(excluding ui and ui+1),should satisfy wx<wui.
Your task is to find the maximum size of Bobo Set in a given tree.

 

Input

The input consists of several tests. For each tests:
The first line contains a integer n (1≤n≤500000). Then following a line contains n integers w1,w2,...,wn (1≤wi≤109,all the wi is distrinct).Each of the following n-1 lines contain 2 integers ai and bi,denoting an edge between vertices ai and bi (1≤ai,bi≤n).
The sum of n is not bigger than 800000.

 

Output

For each test output one line contains a integer,denoting the maximum size of Bobo Set.

 

Sample Input

7
3 30 350 100 200 300 400
1 2
2 3
3 4
4 5
5 6
6 7

 

Sample Output

5

 

/*
 * 在原图中，w[u] < w[v] 则加边 u -> v ， 反之加边 v -> u
 */
#include <bits/stdc++.h>
#define endl "\n"
using namespace std;
typedef long long ll;

const int MAXN = 500000 + 7;
vector<int> linker[MAXN];
int vis[MAXN];
int val[MAXN];
int n, ans = 0;
void Init(){
    for(int i = 0; i <= n; ++i) {
        linker[i].clear();
    }
}

int dfs(int u) {
    if(vis[u]) return vis[u];
    vis[u] = 1;
    for(int i = 0; i < linker[u].size(); ++i) {
        int v = linker[u][i];
        vis[u] += dfs(v);
        ans = max(ans, vis[u]);
    }
    return vis[u];
}

int main() {
    ios::sync_with_stdio(false);
    while(cin >> n) {
        Init();
        for(int i = 1; i <= n; ++i) {
            cin >> val[i];
            vis[i] = 0;
        }
        int a, b;
        for(int i = 1; i < n; ++i) {
            cin >> a >> b;
            if(val[a] < val[b]) {
                linker[a].push_back(b);
            }
            else {
                linker[b].push_back(a);
            }
        }
        ans = 1;
        for(int i = 1; i <= n; ++i) {
            dfs(i);
        }
        cout << ans << endl;
    }
    return 0;
}