poj2342 树形dp

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5844		Accepted: 3354

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 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

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

题意：

某公司要举办一次晚会，但是为了使得晚会的气氛更加活跃，每个参加晚会的人都不希望在晚会中见到他的直接上司，现在已知每个人的活跃指数和上司关系（当然不可能存在环），求邀请哪些人（多少人）来能使得晚会的总活跃指数最大。

思路：

任何一个点的取舍可以看作一种决策，那么状态就是在某个点取的时候或者不取的时候，以他为根的子树能有的最大活跃总值。分别可以用f[i,1]和f[i,0]表示第i个人来和不来。

当i来的时候，dp[i][1] += dp[j][0];//j为i的下属

当i不来的时候，dp[i][0] +=max(dp[j][1],dp[j][0]);//j为i的下属

 

题意：有n个人，接下来n行是n个人的价值，再接下来n行给出l，k说的是l的上司是k，这里注意l与k是不能同时出现的

 

思路：用dp数据来记录价值，开数组用下标记录去或者不去、

 

则状态转移方程为：

 

DP[i][1] += DP[j][0],

 

DP[i][0] += max{DP[j][0],DP[j][1]};其中j为i的孩子节点。

 

这样，从根节点r进行dfs，最后结果为max{DP[r][0],DP[r][1]}。

#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>

using namespace std;

#define maxn 6005

int n;
int dp[maxn][2],father[maxn];//dp[i][0]0表示不去，dp[i][1]1表示去了
bool visited[maxn];

void tree_dp(int node)
{
    int i;
    visited[node] = 1;
    for(i=1; i<=n; i++)
    {
        if(!visited[i]&&father[i] == node)//i为下属
        {
            tree_dp(i);//递归调用孩子结点，从叶子结点开始dp
            //关键
            dp[node][1] += dp[i][0];//上司来,下属不来
            dp[node][0] +=max(dp[i][1],dp[i][0]);//上司不来，下属来、不来
        }
    }
}

int main()
{
    int i;
    int f,c,root;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        memset(father,0,sizeof(father));
        memset(visited,0,sizeof(visited));
        for(i=1; i<=n; i++)
        {
            scanf("%d",&dp[i][1]);
        }
        root = 0;//记录父结点
        bool beg = 1;
        while (scanf("%d %d",&c,&f),c||f)
        {
            father[c] = f;
            if( root == c || beg )
            {
                root = f;
            }
        }
        while(father[root])//查找父结点
            root=father[root];
        tree_dp(root);
        int imax=max(dp[root][0],dp[root][1]);
        printf("%d\n",imax);
    }
    return 0;

}