【树形dp】poj2342

Anniversary party

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

先废话两句：

已经一个月没写题解了真是太过分了……坚决不能再这样了！！！！！！！！！！！

这是一道经典的树形dp，有现成的翻译哈哈：

没有上司的舞会
Ural大学有N个职员，编号为1~N。他们有从属关系，也就是说他们的关系就像一棵以校长为根的树，父结点就是子结点的直接上司。每个职员有一个快乐指数。现在有个周年庆宴会，要求与会职员的快乐指数最大。但是，没有职员愿和直接上司一起与会。
输入格式
 第一行一个整数N。(1<=N<=6000)接下来N行，第i+1行表示i号职员的快乐指数Ri。(-128<=Ri<=127)接下来N-1行，每行输入一对整数L,K。表示K是L的直接上司。最后一行输入0,0。
输出格式
输出最大的快乐指数。
 
刚刚看到这个题的时候还是有点迷茫的，不过想起高中老师说过：树形dp实际上就是在找出转移方程以后后序遍历一次，于是……

f[x][1]表示x要去时能获得的最大的快乐指数

f[x][0]表示x不去时能获得的最大的快乐指数
那么假设k1,k2……ki是x的下属：
f[x][1] =a[x] + f[k1][0] + f[k2][0]+……+ f[ki][0]  //X要去，那他的下属们都不能去
f[x][0] =max(f[k1][0],f[k1][1]) +max(f[k2][0],f[k2][1])+……+ max(f[ki][0]，f[ki][1])
//X不去，那他的下属们可去可不去（取最大）
然后写个后序遍历，然后在返回的时候进行转移就解决问题了
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const long N = 6010;
long n, root;
long a[N], fa[N];
bool v[N];
long f[N][2];
void init()
{
    memset(f, 0, sizeof(f));
    for (long i=1; i<=n; i++)
    {
        scanf("%d", &a[i]);
        f[i][1] = a[i];
    }
    long l, k;
    memset(fa, 0, sizeof(fa));
    while ((scanf("%d%d", &l, &k) != EOF) && (l > 0) && (k > 0))
    {
        fa[l] = k;
    }
    for (long i=1; i<=n; i++)
    {
        if (fa[i] == 0) root = i;
    }


    memset(v, 0, sizeof(v));
}

void dfs(long x)
{

    v[x] = 1;
    for (long i=1; i<=n; i++)
    {
        //printf("%d  %d  %d  %d\n", x, i, v[i], fa[i]);
        if ((!v[i]) && (fa[i] == x))
        {
            dfs(i);
            f[x][0] += max(f[i][1], f[i][0]);
            f[x][1] += f[i][0];
        }
    }

}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        init();
        dfs(root);
        // printf("%d\n", root);
        printf("%d\n", max(f[root][0], f[root][1]));
    }


    return 0 ;
}