Hdu1520 Anniversary party (poj2342, tree dp)

Hdu1520 Anniversary party (poj2342, tree dp)
Anniversary partyTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 7303 Accepted Submission (s): 3220


Problem Description There is going to be a party to celebrate the 80-th Anniversary of the Ural State University. the University has a hierarchical structure of employees. it means that the supervisor relation forms a tree rooted at the rector V. e. tretyakov. in order to make the party funny for every one, the rector does not want both an employee and his or her immediate supervisor to be present. the personnel office has evaluated conviviality of each employee, so everyone has some number (rating) attached to him or her. your task is to make a list of guests with the maximal possible sum of guests 'conviviality ratings.

Input Employees are numbered from 1 to N. A first line of input contains a number N. 1 <= N <= 6 000. each of the subsequent N lines contains the conviviality rating of the corresponding employee. conviviality rating is an integer number in a range from-128 to 127. after that go T lines that describe a supervisor relation tree. each line of the tree specification has the form:
L K
It means that the K-th employee is an immediate supervisor of the L-th employee. Input is ended with the line
0 0
Output shoshould contain the maximal sum of guests ratings.

Sample Input

711111111 32 36 47 44 53 50 0

Sample Output

5

Source Ural State University Internal Contest October '2000 Students Session
This is the same as poj2342, but the hdu data is enhanced. The code using poj2342AC may time out. The method used in the poj2342 parsing report is a bit different, but it is probably the same, it is not difficult to understand that it only adds some parent-child relationship.

# Include
 
  
# Include
  
   
# Include
   
    
# Include
    
     
# Include
     
      
# Include
      # Include
       
         # Include
        
          # Include
         
           # Include using namespace std; const double eps = 1e-6; const double pi = acos (-1.0); const int INF = 0x3f3f3f; const int MOD = 1000000007; # define ll long # define CL (a, B) memset (a, B, sizeof (a) struct Tree {int father; // father int child; // son int brother; // brother int Not; // don't go to party int TakeParty; // go to party int MAX () {return TakeParty> Not? TakeParty: Not;} // The time when the struct function is called. int init () // clear the tree {father = child = brother = Not = 0;} tree [6005]; void dfs (int idx) {int child = tree [idx]. child; while (child) // if there is a child, continue to query {dfs (child); tree [idx]. takeParty + = tree [child]. not; // If idx goes to the party, its child cannot go to the tree [idx]. not + = tree [child]. MAX (); // If idx does not go to the party, its children can go to the instead of the child = tree [child]. brother; // find other children} int main () {int n, a, B; while (scanf (% d, & n) = 1) {for (I Nt I = 1; I <= n; I ++) {scanf (% d, & tree [I]. takeParty); tree [I]. init () ;}while (scanf (% d, & a, & B), a + B) // build {tree [a]. father = B; tree [a]. brother = tree [B]. child; tree [B]. child = a ;}for (int I = 1; I <= n; I ++) // find {if (! Tree [I]. father) {dfs (I); printf (% d, tree [I]. MAX (); break ;}} return 0 ;}