Crazy Bobo

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

Total Submission(s): 1334    Accepted Submission(s): 410

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

 

 解题：直接搜索。。。建立有向图时候，小权向大权的连边，然后看看每个每个点，最多能走多少个点。

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #pragma comment(linker, "/stack:1024000000,1024000000") 7 using namespace std; 8 const int maxn = 500010; 9 vector
           
            g[maxn];10 int w[maxn],ret[maxn];11 void dfs(int u) {12 ret[u] = 1;13 for(int i = g[u].size()-1; i >= 0; --i) {14 if(!ret[g[u][i]]) dfs(g[u][i]);15 ret[u] += ret[g[u][i]];16 }17 }18 int main() {19 int n,u,v;20 while(~scanf("%d",&n)){21 for(int i = 1; i <= n; ++i){22 scanf("%d",w+i);23 g[i].clear();24 }25 for(int i = 1; i < n; ++i){26 scanf("%d%d",&u,&v);27 if(w[u] < w[v]) g[u].push_back(v);28 else g[v].push_back(u);29 }30 memset(ret,0,sizeof ret);31 int ans = 0;32 for(int i = 1; i <= n; ++i){33 if(!ret[i]) dfs(i);34 ans = max(ans,ret[i]);35 }36 printf("%d\n",ans);37 }38 return 0;39 }