Crazy Bobo

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Problem Description

Bobo has a tree,whose vertices are conveniently labeled by 1,2,...,n.Each node has a weight .
All the weights are distrinct.

A set with m nodes  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 ,(that
is, for
i from 1 to m-1).For any node  in
the path from  to (excluding  and ),should
satisfy .

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 ().
Then following a line contains n integers  (,all
the  is
distrinct).Each of the following n-1 lines contain 2 integers  and ,denoting
an edge between vertices  and  ().

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

Source

2015 Multi-University Training Contest 3

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/*

參考此人博客 :http://www.mamicode.com/info-detail-948802.html
记得用c++交

*/

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
using namespace std;

#define N 800005

vector<int>g[N];
int n;
int ans[N];
int a[N];

int dfs(int u)
{
    if(ans[u]) return ans[u];
    ans[u]=1;
    for(int i=0;i<g[u].size();i++)
    {
        int to=g[u][i];
        ans[u]+=dfs(to);
    }
    return ans[u];
}

int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(i=1;i<=n;i++)
            g[i].clear();
        memset(ans,0,sizeof(ans));
        int u,v;
        i=n-1;
        while(i--)
        {
            scanf("%d%d",&u,&v);
            if(a[u]<a[v]) g[u].push_back(v);
            else g[v].push_back(u);
        }
        int temp=0;
        for(i=1;i<=n;i++)
        {
            temp=max(temp,dfs(i));
        }
        printf("%d\n",temp);
    }
    return 0;
}