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;
}
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