POJ 2342 Anniversary party (樹形dp 入門題)

POJ 2342 Anniversary party (樹形dp 入門題)

 

Anniversary party Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4810   Accepted: 2724

 

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

題目鏈接：http://poj.org/problem?id=2342

題目大意：一棵樹，每個節點有一個值，現在要從中選一些點，要求這些點值和最大並且每對兒子和父親不能同時被選

題目分析：dp[i][0]和dp[i][1]分別表示不選和選第i個點時以i為子樹根時子樹值的和，則：
dp[fa[i]][1] += dp[i][0] 表示選i的父親時，其值等於自身值加上不選i時的值
dp[fa[i]][0] += max(dp[i][1], dp[i][0]) 表示不選父親時，則其值等於其兒子被選或沒被選的值的最大值

#include 
#include 
#include 
#include 
using namespace std;
int const MAX = 6005;
int dp[MAX][2], val[MAX];
bool vis[MAX], flag[MAX];
vector  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個點n－1條邊，必然存在“入度”為0的點即樹根
        {
            if(!flag[i])
            {
                root = i;
                break;
            }
        }
        DFS(root);
        printf("%d\n", max(dp[root][1], dp[root][0]));
    }
}