POJ2342 Anniversary party

鏈接：http://poj.org/problem?id=2342

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10326		Accepted: 5913

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

樹形DP：看着像搜索哈。。。

題意：公司人要開party，要從N個人之間選幾人是興奮值最大（2--N+1行，每個人都有唯一的興奮值）

剩下幾行是從屬關係，左邊的是右邊的直接下屬，但選擇的時候，要避開這種直接領導關係，最多只能去一人，問興奮值最大是多少？

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define maxn 7000
using namespace std;
int dp[maxn][2];//1 去 0 不去
int father[maxn];
int vis[maxn];
int n;
void dfs(int root)
{
    for(int i=1;i<=n;i++)
    {
        if(father[i]==root)
        {
            dfs(i);
            dp[root][1]+=dp[i][0];
            dp[root][0]+=max(dp[i][0],dp[i][1]);
        }
    }
}
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&dp[i][1]);
        }
        int l,k;
        int root=0;
        int flag=1;
        memset(vis,0,sizeof(vis));
        while(scanf("%d%d",&l,&k),l||k)
        {
            father[l]=k;
            if(root==l||flag)
            {
                root=k;
            }
        }
        while(father[root])
        {
            root=father[root];
        }
        dfs(root);
        int ans=max(dp[root][0],dp[root][1]);
        printf("%d\n",ans);
    }
    return 0;
}