HDU 5325 Crazy Bobo

HDU 5325 Crazy Bobo

Set the direction of the original edge (u, v) to u-> v when w [u]> w [v]

The maximum Set is the Set of vertices reached by max {u}

 

Crazy Bobo Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission (s): 382 Accepted Submission (s): 116


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 For I from 1 M-1). For any node X In the path from Ui To Ui + 1 (Excluding Ui And Ui + 1 ), Shoshould satisfy Wx .
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 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

73 30 350 100 200 300 4001 22 33 44 55 66 7

Sample Output

5

Source 2015 Multi-University Training Contest 3

#pragma comment(linker, /STACK:1024000000,1024000000)#include 
 
  #include 
  
   #include 
   
    #include using namespace std;typedef long long ll;#define prt(k) cout<<#k = <
    
      w[v]) g[u] = up(v) + 1;        return g[u];}int n;pair
     
       p[N];int main(){        while (scanf(%d, &n)==1) {                for (int i=1;i<=n;i++) scanf(%d, w+i), p[i]=make_pair(w[i], i);                sort(p+1, p+n+1);                mm = 0; memset(head, -1, sizeof head);                for (int i=0;i
      
       =1;i--) { int u = p[i].second; dp[u] = 1; for (int j=head[u];~j;j=e[j].next) { int v = e[j].v; if (w[u] < w[v]) dp[u] += dp[v]; } ans = max(ans, dp[u]); } printf(%d, ans); } return 0;}