HDU

Time limit ：1000 ms ；Memory limit ：32768 kB； OS ：Windows

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

题意：给你一棵树，你从中选择若干结点，要求这些结点中任意两个不能有“父子”关系，求最后权值的最大值。

分析：这便是紫书上所说的“树的最大独立集”问题，这要是放在线性表上，直接用01背包就解决了，但树上的动态规划就是把这种简单操作放在树上解决，难度就增加了。解题步骤如下：

---

①用邻接表储存所给的树。

②因为每个点可取可不取，dp[i][0]表示第i个结点不选时的最大值，dp[i][1]表示第i个结点选时的最大值。状态转移方程为：    dp[father][0] += max(dp[son][0],dp[son][1]);父亲没有被邀请 ，dp[father][1] += dp[son][0];父亲被邀请了，儿子不能被邀请。

③最终结果为max(dp[root][1],dp[root][0])。

---

 AC代码如下（93ms）：

#include <bits/stdc++.h>
using namespace std;
#define fast                                ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll                                  long long
#define _for(i,a,b)                         for(int i = a;i < b;i++)
#define rep(i,a,b)                          for(int i = a;i <= b;i++)
#define all(s)                              s.begin(), s.end()

const int maxn = 6050;
int N;
bool no_root[maxn];//判断是不是树根
vector<int>G[maxn];//邻接表
int dp[maxn][2];//dp[i][0]表示以i为根的子树，不选i时的最大值
				//dp[i][1]表示以i为根的子树，选i时的最大值
void dfs(int root)
{
	int nson = G[root].size();
	_for(i, 0, nson)
	{
		int son = G[root][i];
		dfs(son);
		dp[root][0] += max(dp[son][1], dp[son][0]);
		dp[root][1] += dp[son][0];
	}
}

int main()
{
	while (scanf("%d", &N) == 1)
	{
		memset(no_root, 0, sizeof(no_root));
		rep(i, 1, N)G[i].clear();//清空邻接表
		rep(i, 1, N)
		{
			scanf("%d", &dp[i][1]);
			dp[i][0] = 0;
		}
		int v, u;//员工与直属上级
		while (scanf("%d%d", &v, &u) == 2 && v + u)//u是v的父结点
		{
			G[u].push_back(v);
			no_root[v] = true;//v不是根
		}
		int ans = 0;
		rep(i, 1, N)
		{
			if (no_root[i] == 0)//从根结点开始(注意，这里可能不止一个根)
			{
				dfs(i);
				ans = max(ans, max(dp[i][0], dp[i][1]));
			}
		}
		printf("%d
", ans);
	}
	return 0;
}