hdu5325 树的思维题

http://acm.hdu.edu.cn/showproblem.php?pid=5325

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
  
  


  

/**
hdu5325 树的思维题
题目大意：给定一颗树，每个点都有一个权值，求出一棵子树，权值递增排序，要求相邻权值两个点的路径上的点的权值都要比这两个权值小
解题思路：对于树上的每条边，对于Wu<Wv，连一条u到v的边，从每个点开始bfs，找出能遍历最多的点就是答案
*/
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
const int maxn=1000505;
int head[maxn],ip;
void  init()
{
    memset(head,-1,sizeof(head));
    ip=0;
}

struct note
{
    int v,next;
}edge[maxn];

void addedge(int u,int v)
{
    edge[ip].v=v,edge[ip].next=head[u];head[u]=ip++;
}

pair <int ,int > p[maxn];
int w[maxn],dp[maxn],n;

int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
            p[i]=make_pair(w[i],i);
        }
        sort(p+1,p+n+1);
        init();
        for(int i=1;i<n;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        int ans=0;
        for(int i=n;i>=1;i--)
        {
            int u=p[i].second;
            dp[u]=1;
            for(int j=head[u];j!=-1;j=edge[j].next)
            {
                int v=edge[j].v;
                if(w[u]<w[v])
                {
                    dp[u]+=dp[v];
                }
            }
            ans=max(ans,dp[u]);
        }
        printf("%d\n",ans);
    }
    return 0;
}