POJ2343

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 should 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

树形dp入门题~

/* * 题目意思就是,有一颗关系树,如果第i个人去，那么第i个人的father 就不能去，以第i个人为father的人，也都不能去; * 如果第i个人去，那么第i个人的father可以去或不去，以第i个人为father的人，可以去或不去。 * 那么，可得到dp状态转移方程 * dp[i][0]表示第i个人不去，dp[i][1]表示第i个人去. * 可以由树的根节点递归分析每个节点的孩子节点, * 那么可得到dp[i][0] = Math.max(dp[j][0],dp[j][1]), j is son of i *      和dp[i][1] = Math.max(dp[j][0]), j is son of i */import java.util.Scanner;public class Main {    final static int maxn = 10000+5;    static int[][] dp = new int[maxn][2];    static int[] father = new int[maxn];    static int[] vis = new int[maxn];    static int n;    public static void main(String[] args) {        Scanner in = new Scanner(System.in);        n = in.nextInt();        for (int i = 1; i <= n; i++) {            dp[i][1] = in.nextInt();        }        int a, b, root = 0;        while (in.hasNext()) {            a = in.nextInt();            b = in.nextInt();            if (a == 0 && b == 0) break;            father[a] = b;            root = b;        }        while (father[root] != 0) {            root = father[root];        }        tree_dp(root);        System.out.println(Math.max(dp[root][0], dp[root][1]));    }    private static void tree_dp(int root) {        // TODO Auto-generated method stub        vis[root] = 1;        for (int i = 1; i <= n; i++) {            if (vis[i] == 0 && father[i] == root) {                tree_dp(i);                dp[root][0] += Math.max(dp[i][0], dp[i][1]);                dp[root][1] += dp[i][0];            }        }    }}