hdu 1520 Anniversary party（树形DP）

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ACM 图论——树形DP

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针对一个员工层级结构，设计算法选出参加庆典的员工组合，使总欢乐值最大化，同时确保无上下级同时出席。

题目链接：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): 7637 Accepted Submission(s): 3348

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

Sample Output

Source

Ural State University Internal Contest October'2000 Students Session

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题目大意：n个结点，接下去n行分别表示第i个人具有的活跃值，再输入a，b表示b是a的上司。题目要求具有直接上司和下属关系的两个人不能同时参加party，最终输出让party活跃值最大是多少。

解题思路：用dp[i][j]二维数组来计算，i来表示第几个人，j表示是去或者不去。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。

先得出所以子节点的去或者不去的最大值，通过子节点的值得到当前值。

1、上司参加，直属下属就一定不可以参加。

2、上司不参加，直属下属可以参加也可以不参加。

详见代码。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>

using namespace std;

vector<int>V[6010];
int dp[6010][10];

void Dfs(int root)
{
    int num=V[root].size();//求当前节点下有几个子节点
    for (int i=0; i<num; i++)
    {
        Dfs(V[root][i]);
    }
    for (int i=0; i<num; i++)
    {
        dp[root][0]+=max(dp[V[root][i]][0],dp[V[root][i]][1]);
        dp[root][1]+=dp[V[root][i]][0];
    }
}

int main()
{
    int t,a,b;
    int val[6010];
    int fa[6010];
    while (~scanf("%d",&t))
    {
        memset(dp,0,sizeof(dp));
        for (int i=1; i<=t; i++)
        {
            scanf("%d",&val[i]);
            V[i].clear();
            fa[i]=i;
            dp[i][1]=val[i];
        }
        while (~scanf("%d%d",&a,&b))
        {
            if (a==0&&b==0)
                break;
            fa[a]=b;
            V[b].push_back(a);
        }
        int root=1;
        while (fa[root]!=root)//找顶点
            root=fa[root];
        Dfs(root);
        int ans=max(dp[root][0],dp[root][1]);
        cout<<ans<<endl;
    }
    return 0;
}
</span>