HDU 1520.Anniversary party【树型DP】【8月15】

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9204    Accepted Submission(s): 3952

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

711111111 32 36 47 44 53 50 0

 

Sample Output

5

题意：员工跟领导参加party，每个人都有一个快乐值，当领导去的时候，属于他的员工就不去；领导不去的时候，他的员工可去可不去。求最大快乐只

思路：树型DP。ans[i][0]表示i不去，ans[i][1]表示i去。则ans[i][0] = max(他的员工去的快乐值, 他的员工不去的快乐值)，ans[i][1] = 他的员工不去+他的快乐值。

#include<iostream>#include<cstdio>#include<vector>#include<cstring>using namespace std;const int MAX = 6005;int N;vector <int> v[MAX];int fa[MAX], ha[MAX], a, b, root, ans[MAX][2];int fi(int root){    if(fa[root] == root) return root;    else fa[root] = fi(fa[root]);}void DFS(int root){    ans[root][1] = ha[root];    for(vector<int>::iterator it = v[root].begin(); it != v[root].end(); ++it)    {        DFS(*it);        ans[root][0] += max(ans[*it][1], ans[*it][0]);        ans[root][1] += ans[*it][0];    }}int main(){    while(scanf("%d", &N) != EOF)    {        memset(ha, 0, sizeof(ha));        memset(ans, 0, sizeof(ans));        for(int i = 1; i <= N; ++i)        {            fa[i] = i;            v[i].clear();            scanf("%d", &ha[i]);        }        while(scanf("%d %d", &a, &b) != EOF && a && b)        {            v[b].push_back(a);            fa[a] = fi(b);        }        root = fi(1);        DFS(root);        cout << max(ans[root][0], ans[root][1]) << endl;    }    return 0;}