树形dp

poj 2342

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9480		Accepted: 5454

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

思路分析 ：
　　给出 n 个人， 并且他们之间有一定的上下级关系， 在纸上画一画 会发现她更像一棵树， 但是写此类 dp 就需要你对递归有一个比较好的理解， 模拟建立一棵树， 找到树的根节点，采用 dfs 遍历一遍树，并且将走过的点做一个标记

状态转移方程 ：
　　dp[node][1] += dp[i][0];
　　dp[node][0] += max(dp[i][0], dp[i][1]);

代码示例：

int n;
int val[6005];
int pt[6005];
vector<int>ve[6005];
int dp[6005][2];

void dfs(int x){
    
    for(int i = 0; i < ve[x].size(); i++){
        int to = ve[x][i];
        
        dfs(to);
        dp[x][1] += dp[to][0];
        dp[x][0] += max(dp[to][1], dp[to][0]);
    } 
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int x, y;
    
    cin >> n;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &val[i]);
        dp[i][1] = val[i];
    }
    for(int i = 1; i < n; i++){
        scanf("%d%d", &x, &y);
        ve[y].push_back(x);
        pt[x] = 1;        
    }    
    int root;
    for(int i = 1; i <= n; i++) if (!pt[i]) {root = i; break;}
    dfs(root);
    
    //printf("+++++ %d \n", dp[3][0]);
    printf("%d\n", max(dp[root][0], dp[root][1]));
    return 0;
}