poj 2342 Anniversary party (树形dp）-优快云博客

本文链接：https://blog.youkuaiyun.com/dsaghjkye/article/details/79578256

解决一个关于员工派对的问题，通过算法确保派对上不出现直属上司与下属，同时使得参与者的欢乐值总和最大。采用深度优先搜索和动态规划的方法。

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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

Sample Output

题意：有n个人要开1个party ，他们每个人有一个欢乐值，他们可能会有上司和下属，为了保证玩的开心，party上不能有直属的上司和下属，现在选一些人去参加求最大欢乐总值；

解定义 dp[i][0]为第i个人不参加获得的最大欢乐值 dp[i][1]为第i个人参加获得的最大欢乐值

若j是i的下属

dp[i][0]+=max(dp[j][0],dp[j][1]) +=是因为1个上次可能有多个下属，若i不参加则j可以参加或者j不参加，取最大欢乐值

dp[i][1]+=dp[j][0] i参加 j不能参加

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<stack>
#include<vector>
#include<map>
#define nn 100001
#define mm 500001
#define inff 0x3fffffff
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
int vis[nn];
int dp[nn][2];
int n,num;
int p[nn];
struct node
{
    int en,next;
}e[nn];
int init()
{
    memset(p,-1,sizeof(p));
    memset(dp,0,sizeof(dp));
    memset(vis,0,sizeof(vis));
    num=0;
}
int add(int st,int en)
{
    e[num].en=en;
    e[num].next=p[st];
    p[st]=num++;
}
int dfs(int u,int pre)
{
    vis[u]=1;
    for(int i=p[u];i+1;i=e[i].next)
    {
        int v=e[i].en;
        if(!vis[v])
        {
            dfs(v,u);
            dp[u][1]+=dp[v][0];
            dp[u][0]+=max(dp[v][0],dp[v][1]);
        }
    }
}
int main()
{
    int a,b;
    while(~scanf("%d",&n))
    {
        init();
        for(int i=1;i<=n;i++)
            scanf("%d",&dp[i][1]);
        while(scanf("%d%d",&a,&b)&&a+b)
        {
            add(a,b);
            add(b,a);
        }
        dfs(1,-1);
        printf("%d\n",max(dp[1][0],dp[1][1]));
    }
}