HDU1520: Anniversary party (tree DP)

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
Output shoshould contain the maximal sum of guests ratings.

 

Sample Input
7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0

Sample Output
5

 

However, if you paste the POJ code to WA and add an exit condition where n is 0, it is TLE.

Then I also found that if the function in the structure body is called for 109 ms

If an external function is used, the same timeout occurs.

It seems that calling functions in the structure is time-saving.

# Include <stdio. h> # include <string. h ># include <algorithm> using namespace std; struct node {int child, father, brother, present, not_present; int max () // It takes less time to call a function in the structure body {return present> not_present? Present: not_present;} void init () {child = father = brother = not_present = 0 ;}} tree [6005]; void dfs (int root) {int son = tree [root]. child; while (son) {dfs (son); tree [root]. present + = tree [son]. not_present; tree [root]. not_present + = tree [son]. max (); son = tree [son]. brother ;}} int main () {int n, I, j, k, l; while (~ Scanf ("% d", & n) {for (I = 1; I <= n; I ++) {scanf ("% d ", & tree [I]. present); tree [I]. init ();} while (~ Scanf ("% d", & l, & k), l + k) {tree [l]. father = k; tree [l]. brother = tree [k]. child; tree [k]. child = l ;}for (I = 1; I <= n; I ++) {if (! Tree [I]. father) {dfs (I); printf ("% d \ n", tree [I]. max (); break ;}} return 0 ;}