hdu 1520 Anniversary party 基础树dp

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6058 Accepted Submission(s): 2743

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

1 3

2 3

6 4

7 4

4 5

3 5

0 0

Sample Output

Source

Ural State University Internal Contest October'2000 Students Session

思路：dp[i][0]， dp[i][1]分别表示不取节点 i 上的值和取节点 i 上的值后，以 i 为根的树在满足题目要求的下能得到的最大值

叶子节点：dp[i][0] = 0, dp[i][1] = v[i] ;

非叶子节点：　　　dp[i][0] = sum( max( dp[j][0], dp[j][1] ) ) , dp[i][1] = sum( dp[j][0] ) + v[i] ;

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstdlib>
 4 #include <algorithm>
 5 #include <cstring>
 6 #include <vector>
 7 #include <map>
 8 using namespace std ;
 9 const int N = 6050 ;
10 int v[N], dp[N][2] ;
11 vector<int> G[N] ;
12 int n, root, vis[N] ;
13 void _in()
14 {
15     memset(vis, 0, sizeof vis) ;
16     for(int i = 0; i <= n; ++i) G[i].clear() ;
17     for(int i = 1; i <= n; ++i) scanf("%d",&v[i]) ;
18     int u, v ;
19     while(1){
20         scanf("%d%d",&v,&u) ;
21         if(!v && !u) break ;
22         vis[v] = 1 ;
23         G[u].push_back(v) ;
24     }
25     for(int i = 1; i <= n; ++i)
26     if(!vis[i]) { root = i ; break; }
27     //printf("%d--\n",root) ;
28 }
29 int get(int root, int c)
30 {
31     int& res = dp[root][c] ;
32     if(res != -1) return res ;
33     int sx = G[root].size() ;
34     if(c) res = v[root] ;
35     else  res = 0 ;
36     for(int i = 0; i < sx; ++i)
37     if(c) res += get(G[root][i],0) ;
38     else res += max(get(G[root][i],0), get(G[root][i],1)) ;
39     return res ;
40 }
41 int main()
42 {
43     #ifdef LOCAL
44     freopen("in.txt","r",stdin) ;
45     #endif
46     while(~scanf("%d",&n)){ //多case，囧
47     _in() ;
48     memset(dp, -1, sizeof dp) ;
49     printf("%d\n",max(get(root,0), get(root,1))) ;
50     }
51     return 0 ;
52 
53 }