多校 hdu 5325

Crazy Bobo

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1308    Accepted Submission(s): 400

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
  

 

Author

ZSTU

 

Source

2015 Multi-University Training Contest 3

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define INF 0x3f3f3f3f
#define LL long long
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))

#define maxn 500000 + 10

int w[maxn], a[maxn];

struct edge
{
    int v, next;
}e[maxn * 2];

int n;
int vis[maxn];
int head[maxn], tot;

void init()
{
    tot = 0;
    memset(vis, 0, sizeof(vis));
    memset(a, 0, sizeof(a));
    memset(head, -1, sizeof(head));
}

void adde(int u, int v)
{
    e[tot].v = v;
    e[tot].next = head[u];
    head[u] = tot++;
}

void dfs(int u)
{
    vis[u] = 1;
    a[u] = 1;
    for(int i=head[u]; i!=-1; i=e[i].next)
    {
        int v = e[i].v;
        //printf("%d--%d\n", u, v);
        if(!vis[v])
            dfs(v);
        a[u] += a[v];
    }
}

int main()
{
    while(~scanf("%d", &n))
    {
        init();
        for(int i=1; i<=n; i++)
            scanf("%d", w + i);
        for(int i=1; i<n; i++)
        {
            int u, v;
            scanf("%d%d", &u, &v);
            if(w[u] < w[v])
                adde(u, v);
            else if(w[u] > w[v])
                adde(v, u);
        }
        for(int i=1; i<=n; i++)
            if(!vis[i])
            dfs(i);

        int ans = -1;
        for(int i=1; i<=n; i++)
            ans = max(ans, a[i]);

        printf("%d\n", ans);
    }
    return 0;
}