hdu 1520 Anniversary party(入门树形DP)

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6926		Accepted: 3985

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

Source

Ural State University Internal Contest October'2000 Students Session

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
#include<algorithm>
#define N 6050
using namespace std;
vector<int >v[N];//用来建树
int w[N];//用来存放每个点的权值
int f[N];//用来找根节点在哪的
int dp[N][2];//dp[i][j]表示第i个人来或不来的最大权值
void dfs(int root)
{
    int len=v[root].size();//这个结点的儿子数
    dp[root][1]=w[root];
    for(int i=0;i<len;i++)
        dfs(v[root][i]);//递归遍历下一层的儿子
    for(int i=0;i<len;i++)
    {
        dp[root][0]+=max(dp[v[root][i]][1],dp[v[root][i]][0]);
        dp[root][1]+=dp[v[root][i]][0];
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        //cout<<n<<endl;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&w[i]);
            dp[i][1]=dp[i][0]=0;//这个人来或不来要清零
            v[i].clear();
            f[i]=-1;
        }
        int a,b;
        while(scanf("%d%d",&a,&b)!=EOF)
        {
            //cout<<a<<" "<<b<<endl;
            if(a==0&&b==0) break;
            f[a]=b;
            v[b].push_back(a);//b结点有儿子a
        }    
        int root=1;
        while(f[root]!=-1) 
        {
            //cout<<"root="<<root<<endl;
            root=f[root];//找树根
        }    
        //cout<<"root="<<root<<endl;
        dfs(root);
        printf("%d\n",max(dp[root][1],dp[root][0]));
    }
    return 0;
}