POJ2342: Anniversary party (tree DP)

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 N-1 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. n people, n rows are the value of n people, and n rows are the value of l. k indicates that l's boss is k, note that l and k cannot appear at the same time: Use dp data to record value, and use subscript to record or not to open an array, the state transition equation is: DP [I] [1] + = DP [j] [0], DP [I] [0] + = max {DP [j] [0], DP [j] [1]}, where j is the child node of I. In this way, dfs is performed from the root node r, and the final result is max {DP [r] [0], DP [r] [1]}.

# Include <stdio. h> # include <string. h ># include <algorithm> using namespace std; int father [6005], vis [6005], dp [6005] [2], t; void dfs (int node) {int I, j; vis [node] = 1; for (I = 1; I <= t; I ++) {if (! Vis [I] & father [I] = node) {dfs (I); dp [node] [1] + = dp [I] [0]; // If node is removed, I cannot go to dp [node] [0] + = max (dp [I] [0], dp [I] [1]); // node does not go, take the maximum value of I or does not go} int main () {int I, j, l, k, root; while (~ Scanf ("% d", & t) {for (I = 1; I <= t; I ++) scanf ("% d ", & dp [I] [1]); root = 0; while (scanf ("% d", & l, & k), l + k> 0) {father [l] = k; // record supervisor root = k;} memset (vis, 0, sizeof (vis); dfs (root ); printf ("% d \ n", max (dp [root] [1], dp [root] [0]);} return 0 ;}