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
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
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 }