Anniversary party HDU - 1520（樹形dp）

傳送門

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 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      

題意：要求從n個人中選出一些人來參加party，這n個人不能有直接的上下級關系，每個人有他的價值，要求最後獲得的最大的總價值是多少？

基礎的樹形dp吧，我們用dp[i][0]表示i沒有來，用dp[i][1]表示i過來了，那麼狀态轉移方程也是很明顯的，首先上機若果來了，則他的下級一定不會過來，反之，他的下級可以過來或者不過來。

dp[fa][0] += max(dp[G[fa][i]][1] , dp[G[fa][i]][0]);

dp[fa][1] += dp[G[fa][i]][0];

#include<iostream>
#include<stdio.h>
#include<string>
#include<math.h>
#include<queue>
#include<vector>
#include<string.h>
#include<iterator>
using namespace std;
const int inf = 0x3f3f3f3f;
int val[6005];
int fa[6005];
int dp[6005][2];
vector<int> G[6005];
void dfs(int fa) {
    for(int i = 0 ; i < G[fa].size() ; i ++) {
        int son = G[fa][i] ;
        dfs(son) ;
    }
    for(int i = 0 ; i < G[fa].size() ; i ++) {
        dp[fa][0] += max(dp[G[fa][i]][1] , dp[G[fa][i]][0]);
        dp[fa][1] += dp[G[fa][i]][0];
    }
}
int main()
{
    int n ;
    while(~scanf("%d" , &n)) {
        memset(fa , 0 , sizeof(fa));
        memset(dp , 0 , sizeof(dp));
        for(int i = 0 ; i <= n ; i ++)
            G[i].clear();
        for(int i = 1 ; i <= n ; i ++) {
            scanf("%d" , &val[i]);
            fa[i] = i;
            dp[i][1] = val[i];
        }
        int u , v ;
        while(1){
            scanf("%d%d" , &u , &v) ;
            if(u == 0 && v == 0)
                break;
            G[v].push_back(u);
            fa[u] = v ;
        }
        int root = 1;
        while(fa[root] != root)
            root = fa[root] ;
        dfs(root);
        cout<<max(dp[root][0] , dp[root][1])<<endl;
    }
    return 0;
}