HDU1520 Anniversary party 樹形dp

題目鏈接

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23856    Accepted Submission(s): 8418

 

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

 

有根樹上每個節點有一個權值，父節點與子節點只能選一個，問最大點權和。

樹形dp基礎題。

dp[u][1]表示選擇u節點後子樹最大點權和，dp[u][0]表示不選u後子樹最大點權和。可以知道選父親後兒子節點一定不能選，不選父親時兒子可選可不選。

可以寫出轉移方程

v是u的子節點

dp[u][1]+=dp[v][0];

dp[u][0]+=max(dp[v][1],dp[v][0]);

 

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;
const int N=6e4+10;
int n,dp[N][2],v[N];
bool vis[N];
vector<int>tr[N];
void dfs(int u)
{
	dp[u][1]=v[u];
	dp[u][0]=0;//初始化
	for(int i=0;i<tr[u].size();i++)
	{
		int v=tr[u][i];
		dfs(v);
		dp[u][1]+=dp[v][0];
		dp[u][0]+=max(dp[v][0],dp[v][1]);
	}
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	while(cin>>n)
	{
		int l,k;
		for(int i=1;i<=n;i++)
		{
			cin>>v[i];
			tr[i].clear();
		}
		ms(vis,0);
		while(cin>>l>>k,l+k)
		{	
			tr[k].push_back(l);
			vis[l]=1;
		}
		for(int i=1;i<=n;i++)
			if(!vis[i])
			{
				dfs(i);
				cout<<max(dp[i][0],dp[i][1])<<endl;
				break;
			}
	}
}