HDU-1520 Anniversary party (树形DP）

Problem 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 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

思路：

• 初探树形DP,dp[i][1]代表i到场的最大活跃度，dp[i][0]代表i不到场的最大活跃度。
• 对于每一个结点如下考虑：

1. 如果这个人来则最大活跃度等于自己的活跃度加上所有子节点不来的最大活跃度（子树之间互不影响）；
2. 如果这个人不来的话，其最大活跃度就等于0加上所有子节点的max(来的最大活跃度，不来的最大活跃度）；

• 将所有dp[i][1]初始化为i的活跃度，则有状态转移方程(i为j父节点）：

dp[i][1] += dp[j][0];

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

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 6000 + 10;
 4
 5 vector<int> T[MAXN];
 6 int dp[MAXN][2];
 7
 8 void dfs(int a, int fa) {
 9     int len = T[a].size();
10     for (int i = 0; i < len; ++i)
11         dfs(T[a][i], a);
12     dp[fa][1] += dp[a][0];
13     dp[fa][0] += max(dp[a][0], dp[a][1]);
14 }
15
16 int main() {
17     int n, L, K;
18     while(~scanf("%d", &n)) {
19         memset(dp, 0, sizeof(dp));
20         int root = n * (n + 1) / 2;
21         for (int i = 1; i <= n; ++i) scanf("%d", &dp[i][1]);
22         while(scanf("%d%d", &L, &K), L+K) {
23             T[K].push_back(L);
24             root -= L;
25         }
26         dfs(root, 0);
27         int ans = max(dp[root][1], dp[root][0]);
28         printf("%d\n", ans);
29         for (int i = 1; i <= n; ++i) T[i].clear();
30     }
31     return 0;
32 }