hdu 1520-树形DP

A - Anniversary party

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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
      
      
      
      

 

关于输入输出：

输入： 

输入n个结点，接下去的n行，表示1-n的每个结点分别具有的活跃值，在接下来去的n-1行，输入a,b，表示b是a的上司

输出：

由于直接有上司和下属关系的两个人不能同时参加party， 求出能让party活跃值最大的方案（求出最大的活跃值即可）.

题目的图形：

解题思路：

每个结点有两种状态，参加和不参加，用0表示不参加，1表示参加

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。

状态转移方程：dp[u][0] += max (dp[v][0], dp[v][1]) :表示上司没参加，其员工可以参加可以不参加

dp[u][1] += dp[v][0] : 表示若上司参加了，其员工一定不会参加

树形dp建树：

用连接表，这是一个有向树

代码：

#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <iostream>

using namespace std;

#define N 6005
int dp[N][2], val[N], visit[N];
int in[N];
struct node {
	int v;
	int next;
}pe[N<<1];

int e, head[N];
void init () {
	memset  (head, -1, sizeof (head));
	e = 0;
}
void addedge (int u, int v) {
	pe[e].v = v;
	pe[e].next = head[u];
	head[u] = e ++;
}
int dfs (int u) {
	visit[u] = 1;
	dp[u][0] = 0;
	dp[u][1] = val[u];
	for (int i = head[u]; i != -1; i = pe[i].next) {
		int v = pe[i].v;
		dfs(v);
		dp[u][0] += max(dp[v][0],  dp[v][1]);
		dp[u][1] += dp[v][0];
	}
	return max(dp[u][0], dp[u][1]);
}

int main() {
	int n, v, u;
	while (scanf ("%d", &n) != EOF) {
		init();
		for (int i = 1; i <= n; ++i) {
			in[i] = 0;
			visit[i] = 0;
			scanf ("%d", &val[i]);
		}
		while (scanf ("%d%d", &v, &u) && v + u != 0) {
			addedge (u, v);
			in[v]++;
		}
		int ans = 0;
		for (int i = 1; i <= n; ++i) {
			if(!in[i])
				ans += dfs(i);
		}
		printf("%d\n", ans);
	}
	return 0;
}