ural 1039 Anniversary Party

1039. Anniversary Party

Time limit: 0.5 second
Memory limit: 8 MB

Background

The president of the Ural State University is going to make an 80'th Anniversary party. The university has a hierarchical structure of employees; that is, the supervisor relation forms a tree rooted at the president. Employees are numbered by integer numbers in a range from 1 to N, The personnel office has ranked each employee with a conviviality rating. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend.

Problem

Your task is to make up a guest list with the maximal conviviality rating of the guests.

Input

The first line of the input contains a number N. 1 ≤ N ≤ 6000. 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 the supervisor relation tree goes. Each line of the tree specification has the form

<L> <K>

which means that the K-th employee is an immediate supervisor of L-th employee. Input is ended with the line

0 0

Output

The output should contain the maximal total rating of the guests.

Sample

input	output
7 1 1 1 1 1 1 1 1 3 2 3 6 4 7 4 4 5 3 5 0 0	5

没有上司的年会。ural要举办场年会，请你邀请来宾。要求每位客人的直接上司不被邀请，否则会很尴尬。每位客人有一个兴趣值。要求使来宾的兴趣值之和最大。

每个人可以选择去或者不去。所以：

dp[t][0] 表示以t为最高上司的子树不安排t参加的最大兴趣值。

dp[t][1] 表示以t为最高上司的子树安排t参加的最大兴趣值。

不安排t参加的话，其直系下属可以去或者不去。dp[t][0] = sum(max(dp[ti][0], dp[ti][1]));

安排t参加的话，其直系下属均不能参加。dp[t][1] = sum(dp[ti][0]) + c[t];

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;

int n;
int c[6006];
std::vector<int > v[6006];
int dp[6006][2];

void Tdp(int t) {
    if(v[t].size() == 0) {
        dp[t][0] = 0;
        dp[t][1] = c[t];
        return ;
    }
    for (int i=0; i<v[t].size(); i++) {
        Tdp(v[t][i]);
    }
    for (int i=0; i<v[t].size(); i++) {
        dp[t][0] += max(dp[v[t][i]][0], dp[v[t][i]][1]);
        dp[t][1] += dp[v[t][i]][0];
    }
    dp[t][1] += c[t];
}

int main () {
    cin >> n;
    for (int i=1; i<=n; i++) {
        cin >> c[i] ;
    }
    int l ,k;
    int vis[6006];
    memset(vis, false, sizeof(vis));
    while (cin >> l >> k) {
        if (l == 0 && k == 0) break;
        v[k].push_back(l);
        vis[l] = true;
    }
    int root;
    for (int i=1; i<=n; i++) {
        if (!vis[i]) {
            root = i;//根节点(最高上司)为没有父节点的点。
            break;
        }
    }
    Tdp(root);
    cout << max(dp[root][0], dp[root][1]) <<endl;
    return 0;
}