题目

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 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

题解

树形dp 的样例题

树形dp根据名字可以看出是 树 和 动态规划 的结合
树有递归性,有向性这些特点
动态规划有递推性的特点
树形dp就是他们的结合

如图是样例构成的树

根据题意,求取最大结点和,保证每个结点和他的父节点不同时存在

用 dp[i][1] 表示选取第 i 个结点所能达到的以其为根的树的最大值
用 dp[i][0] 表示不选取第 i 个结点所能达到的以其为根的树的最大值

则,有递推方程
dp[i][0] += dp[j][1]
dp[i][1] += max(dp[j][1],dp[j][0])
其中 j 是 i 的子结点

递归计算即可

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

const int maxn = 6005;

int happy[maxn];
int father[maxn];
vector<int> children[maxn];

int dp[maxn][2];

void tree(int t) {
    for(size_t i = 0;i < children[t].size();i++) {
        int &child = children[t][i];
        tree(child);
        dp[t][1] += dp[child][0];
        dp[t][0] += max(dp[child][1],dp[child][0]);
    }
}

bool Do() {
    int n;
    if(!(cin >> n))
        return false;

    memset(dp,0,sizeof(dp));

    for(int i = 1;i <= n;i++) {
        cin >> dp[i][1];
        children[i].clear();
    }

    while(1) {
        int a,b;
        cin >> a >> b;
        if(a == 0 && b == 0)
            break;
        father[a] = b;
        children[b].push_back(a);
    }

    int root = 1;
    while(father[root])
        root = father[root];

    tree(root);
    cout << max(dp[root][0],dp[root][1]) << endl;
    return true;
}

int main() {
    while(Do());
    return 0;
}