hdu1520:Anniversary party

最新推荐文章于 2024-08-17 03:26:58 发布

原创最新推荐文章于 2024-08-17 03:26:58 发布 · 189 阅读

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本文介绍了一种使用树形动态规划解决特定问题的方法：如何在考虑组织结构限制的前提下，选择一组员工参加聚会以最大化整体亲和力评分。文章详细解释了状态转移方程，并提供了实现代码。

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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

题目大意
给你一棵树以及每个节点上的权值，在父节点与子节点不同时存在的情况下求选出所有节点的一部分，使得节点之和最大。

解题思路
一个普通的树型dp问题，按从叶到根的方向遍历。状态转移方程为dp[i][0]+=max(dp[j][0],dp[j][1])，dp[i][1]+=dp[j][0](j为i的子节点）。

代码

#include<iostream>
#include<memory.h>
#include<algorithm>
#include<vector>
using namespace std;
int n;
int dp[6005][2];
int father[6005];
vector<int>G[6005];
void dfs(int root)
{
    for(int i=0;i<G[root].size();i++)
    dfs(G[root][i]);//从根遍历到叶子结点

    //回溯时更新
    for(int i=0;i<G[root].size();i++)
    {
        dp[root][0]+=max(dp[G[root][i]][0],dp[G[root][i]][1]);
        dp[root][1]+=dp[G[root][i]][0];
    }
}
int main()
{
    int u,v;
    while(cin>>n)
    {
        memset(father,-1,sizeof(father));
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            cin>>dp[i][1];
            G[i].clear();
        }
        while(cin>>u>>v)
        {
            if(u==0&&v==0)
            break;
            father[u]=v;
            G[v].push_back(u);
        }
        int root=1;
        while(father[root]!=-1)
        root=father[root];

        dfs(root);

        cout<<max(dp[root][1],dp[root][0])<<endl;
    }
    return 0;
}