Anniversary party
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4310 Accepted Submission(s): 1976
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
Source
Ural State University Internal Contest October'2000 Students Session
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題目大意：給你一顆樹，是人際關系樹，然後互為上下級的（父子節點）的不能同時選中。
解題思路：樹形dp，隨意從一個點開始擴展，把周圍所有節點的dp都解決出來，然後加上去即可。
用dp[i][0]表示不選中i號，周圍的都可以選的最大值，dp[i][1]表示選中i號，那麼周圍都不能選的最大值。
則
dp[i][0]+=(dp[j][1],dp[j][0])
dp[i][1]+=dp[j][0]
i和j相鄰，並且在求i的時候從j擴展的都已經求出來。
詳見代碼：
題目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1520
AC代碼：
#include
#include
#include
#include
#include
using namespace std;
const int maxn=6005;
vector mq[maxn];
int dp[maxn][2];
void dfs(int cur,int p)
{
int i,nex;
for(i=0;i
