poj 2342 Anniversary party （树形dp入门）

题意：

某公司，每个人都有一个直接上司（除boss外且不存在环），每个人都有一个活跃值。现要举行一个晚会，参加的人中不能有直接上司存在，问最大活跃值是多少？

解题思路：

任何一个人的来或不来可以看作一种决策，那么状态就是在某个人来的时候或者不来的时候，以他为根的子树能有的最大活跃总值。分别可以用dp[i][1]和dp[i][0]表示第i个人来和不来。

所以可以得到状态转移方程：

第i个人来：dp[i][1] += dp[child][0](他的直接下属一定不能来)

第i个人不来：dp[i][0] += max ( dp[child][0], dp[child][1])(他的直接下属可来可不来，取较大值)

代码：

#include<iostream>#include<cstdio>#include<vector>#include<cstring>using namespace std;const int maxn = 1e4+5;vector<int> m[maxn];int n, dp[maxn][2], book[maxn];void dfs(int r){    for(int i = 0; i < m[r].size(); i++)    {        int child = m[r][i];        dfs(child);        dp[r][1] += dp[child][0];        dp[r][0] += max(dp[child][0], dp[child][1]);    }}int main(void){    while(cin >> n)    {        memset(dp, 0, sizeof(dp));        for(int i = 0; i < maxn; i++)            m[i].clear();        for(int i = 1; i <= n; i++)            scanf("%d", &dp[i][1]);        int x, y;        for(int i = 2; i <= n; i++)        {            scanf("%d%d", &x, &y);            book[x] = 1;            m[y].push_back(x);        }        scanf("%d%d", &x, &y);        int root;        for(int i = 1; i <= n; i++)            if(!book[i])            {                root = i;                break;            }        dfs(root);        printf("%d\n", max(dp[root][0], dp[root][1]));    }    return 0;}

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

711111111 32 36 47 44 53 50 0

Sample Output

5