Anniversary party（树形dp入门）

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3043    Accepted Submission(s): 1374

Problem 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 T 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

 

 

Recommend

linle

 

思路：当前节点为i，dp[i][0]为i不出席时的最大快乐值，dp[i][1]为i出席时的最大快乐值。i的儿子集合（直接下属集合）为son={s1，s2，···}，则有

（1）求解dp[i][0]：

for each si in son

    dp[i][0] += max(dp[si][0], dp[si][1]);

解释：i不出席时，i的儿子既可以出席也可以不出席，取快乐值最大的方案。

（2）求解dp[i][1]:

for each si in son

    dp[i][1] += dp[i][0];

解释：i出席，则i的所有儿子只能不出席。

 

AC Code：

#include <iostream>
#include <vector>
#include <cstdio>
using namespace std;

const int MAXN = 6002;
struct maxVal
{
    int y, x;
}dp[MAXN];
//x-不包括自身时最大快乐值；y-包括自身时最大快乐值
vector<int> gra[MAXN];  //邻接表
bool isRoot[MAXN];  //标记是否根节点

int Max(const int& a, const int& b)
{
    return a > b ? a : b;
}

maxVal getMax(int v)  //a-不包括自己，b-包括自己
{
    if(gra[v].empty())
    {
        return dp[v];
    }
    vector<int>::iterator it = gra[v].begin();
    for(; it != gra[v].end(); it++)
    {
        maxVal val = getMax(*it);
        dp[v].x += Max(val.x, val.y);
        dp[v].y += val.x;
    }
    return dp[v];
}

int main()
{
    int n, root;  //root-根节点
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%d", &dp[i].y);
            dp[i].x = 0;
            isRoot[i] = true;
            gra[i].clear();
        }
        int l, k;
        while(scanf("%d %d", &l, &k) && l)
        {
            gra[k].push_back(l);
            isRoot[l] = false;
        }
        for(int i = 1; i <= n; i++)
            if(isRoot[i] == true)
            {
                root = i;
                break;
            }
        maxVal ans = getMax(root);
        printf("%d\n", Max(ans.x, ans.y));
    }
    return 0;
}