传送门：

http://acm.hdu.edu.cn/showproblem.php?pid=1520

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16376    Accepted Submission(s): 6241

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

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题目大意：

学校要开一个聚会。学校的教职工之间有上下级关系，为了让所有人开心，宴会组织者决定不会同时邀请一个人和他的上级（这让我想起我们昨天晚上聚餐李晔老师不来，她怕她来了我们放不开。。。。），对于每一个人，他能给聚会带来的欢乐度有一个值，问组织者该邀请哪些人能够使宴会的欢乐度达到最大值。

解题思路：

首先是DP的部分（也是很无聊的一部分）：每个参与者都有两种状态，一种是参加，一种是不参加。这个状态的后续影响就是如果他参加了，他的直接上司和直接下属都不能参加。我们可以用一个二维二态的数组来描述：dp[i][1]表示第i个参与者参加了，dp[i][0]表示第i个参与者没有参加。状态转移方程就是dp[i][1]=dp[i][1]+dp[i-1][0]，dp[i][0]=dp[i][0]+Max(dp[i-1][0],dp[i-1][1])。

本质：将一个数字三角形的操作放在了树上

第一道树形dp题，纪念一下，虽然对树的部分理解的还不是很透彻

code：

#include <iostream>
#include <stdio.h>
#include<memory>
#include<stack>
#include<string.h>
#include<algorithm>
using namespace std;
#define max_v 6005
struct node
{
    int pa,son;
    int next;
} point[max_v];

int dp[max_v][2];
//dp[i][1]表示第i个参与者参加了，dp[i][0]表示第i个参与者没有参加
//状态转移方程就：
//dp[i][1]=dp[i][1]+dp[i-1][0]
//dp[i][0]=dp[i][0]+Max(dp[i-1][0],dp[i-1][1])

int List[max_v];

int vis[max_v];//vis[a]=1 表示a有父节点

int value[max_v];//存值
int pos;

void add(int pa,int son)
{
    point[pos].pa=pa;
    point[pos].son=son;
    point[pos].next=List[pa];
    List[pa]=pos++;
}

void dfs(int root)
{
    if(List[root]==-1)//root没有子节点了
    {
        dp[root][1]=value[root];
        dp[root][0]=0;
        return ;
    }

    int now=List[root];
    dp[root][0]=0;
    dp[root][1]=value[root];

    while(now!=-1)
    {
        dfs(point[now].son);
        dp[root][1]+=dp[point[now].son][0];//既然取了父节点的值，子节点的值就不能再取了。

        //父节点的值没有取，子节点的值分取和不取两种情况，取其中较大的那种情况。
        dp[root][0]+=max(dp[point[now].son][1],dp[point[now].son][0]);

        now=point[now].next;//这个子节点计算过了，就要开始计算下一个子节点了
    }
    return ;
}

int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1; i<=n; i++)
            cin>>value[i];//记录每一个点的值

        memset(List,-1,sizeof(List));
        memset(vis,0,sizeof(vis));

        int a,b;
        pos=0;

        while(~scanf("%d %d",&a,&b))
        {
            if(a==0&&b==0)
                break;
            add(b,a);  //将边加入树中
            vis[a]=1; //记录a有父节点，不可能是祖节点。
        }

        a=1;
        while(vis[a]==1)
            a++;//找到根结点

        dfs(a);//从根结点开始搜

        printf("%d\n",max(dp[a][0],dp[a][1]));//取最大
    }
    return 0;
}

原文地址：https://www.cnblogs.com/yinbiao/p/9406533.html