多校3 1010 Crazy Bobo

Crazy Bobo

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 879    Accepted Submission(s): 264

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

 

Source

2015 Multi-University Training Contest 3

 

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#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;

vector<int> edg[500005];
struct Node
{
    int w;
    int num;
}a[500050];

int n;

bool cmp(Node i,Node o)
{
    return i.w>o.w;
}

int dp[500005];

int main()
{
    int i,j,k;
    int x,y;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
            edg[i].clear();
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i].w);
            a[i].num=i;
        }
        for(i=1;i<n;i++)
        {
            scanf("%d %d",&x,&y);
            if(a[x].w<a[y].w)
                edg[x].push_back(y);
            else
                edg[y].push_back(x);
        }
        sort(a+1,a+n+1,cmp);
        int ma=1;
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            int u=a[i].num;
            dp[u]=1;
            for(j=0;j<edg[u].size();j++)
            {
                int v=edg[u][j];
                dp[u]=dp[u]+dp[v];
            }
            if(dp[u]>ma)
                ma=dp[u];
        }
        printf("%d
",ma);

        /*printf("***
");
        for(i=1;i<=n;i++)
            printf("%d ",a[i].num);
        printf("
");
        for(i=1;i<=n;i++)
            printf("%d ",dp[a[i].num]);
        printf("
");
        for(j=0;j<edg[6].size();j++)
            printf("%d ",edg[6][j]);
        printf("
");*/
    }
    return 0;
}