POJ 2342 Anniversary party (树形dp 入门题)

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

Source

Ural State University Internal Contest
October‘2000 Students Session

题目链接：http://poj.org/problem?id=2342

题目大意：一棵树，每个节点有一个值，现在要从中选一些点，要求这些点值和最大并且每对儿子和父亲不能同时被选

题目分析：dp[i][0]和dp[i][1]分别表示不选和选第i个点时以i为子树根时子树值的和，则：

dp[fa[i]][1] += dp[i][0]  表示选i的父亲时，其值等于自身值加上不选i时的值

dp[fa[i]][0] += max(dp[i][1], dp[i][0])  表示不选父亲时，则其值等于其儿子被选或没被选的值的最大值

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
using namespace std;
int const MAX = 6005;
int dp[MAX][2], val[MAX];
bool vis[MAX], flag[MAX];
vector <int> vt[MAX];

void DFS(int fa)
{
    vis[fa] = true;
    dp[fa][1] = val[fa];
    int sz = vt[fa].size();
    for(int i = 0; i < sz; i++)
    {
        int son = vt[fa][i];
        if(!vis[son])
        {
            DFS(son);
            dp[fa][1] += dp[son][0];
            dp[fa][0] += max(dp[son][1], dp[son][0]);
        }
    }
    return;
}

int main()
{
    int n;
    while(scanf("%d", &n) && n)
    {
        for(int i = 1; i <= n; i++)
            vt[i].clear();
        memset(flag, false, sizeof(flag));
        memset(vis, false, sizeof(vis));
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
            scanf("%d", &val[i]);
        for(int i = 0; i < n - 1; i++)
        {
            int a, b;
            scanf("%d %d", &a, &b);
            vt[b].push_back(a);
            flag[a] = true;
        }
        int root;
        for(int i = 1; i <= n; i++) //n个点n－1条边，必然存在“入度”为0的点即树根
        {
            if(!flag[i])
            {
                root = i;
                break;
            }
        }
        DFS(root);
        printf("%d\n", max(dp[root][1], dp[root][0]));
    }
}