Anniversary party(hdu1520)

Anniversary party

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9976 Accepted Submission(s): 4234

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

5

思路：树形dp;

dp[i][j] 表示第j个节点状态为i时i ->>[0,1]该节点选或不选，然后从根节点dfs dp就可以了。复杂度O(n);

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<queue>
 4 #include<algorithm>
 5 #include<string.h>
 6 #include<iostream>
 7 #include<stack>
 8 #include<vector>
 9 using namespace std;
10 typedef long long LL;
11 int val[7000];
12 vector<int>vec[7000];
13 int dp[2][7000];
14 int cnt[7000];
15 void dfs(int n);
16 int main(void)
17 {
18     int n;
19     while(scanf("%d",&n)!=EOF)
20     {
21         int i,j;
22         memset(cnt,0,sizeof(cnt));
23         for(i = 0; i < 7000; i++)
24             vec[i].clear();
25         memset(dp,0,sizeof(dp));
26         for(i = 1; i <= n; i++)
27             scanf("%d",&val[i]);
28         int ch,fa;
29         while(scanf("%d %d",&ch,&fa),ch!=0&&fa!=0)
30         {
31             cnt[ch] = 1;
32             vec[fa].push_back(ch);
33         }
34         int root;
35         for(i = 1; i <= n; i++)
36         {
37             if(!cnt[i])
38             {
39                 root = i;
40                 break;
41             }
42         }
43         dfs(root);
44         if(dp[1][root]<0&&dp[0][root]<0)
45         {
46             printf("0
");
47         }
48         else
49         {
50             int ask = max(dp[1][root],dp[0][root]);
51             printf("%d
",ask);
52         }
53     }
54     return 0;
55 }
56 void dfs(int n)
57 {
58     int sum1 = 0;
59     int sum2 = 0;
60     for(int i = 0; i < vec[n].size(); i++)
61     {
62         int id = vec[n][i];
63         dfs(id);
64         sum1 += dp[0][id];
65         sum2 += max(max(dp[1][id],dp[0][id]),0);
66     }
67     dp[0][n] = max(dp[0][n],sum2);
68     dp[0][n] = max(dp[0][n],sum1);
69     dp[1][n] = max(dp[0][n],val[n]);
70     dp[1][n] = max(dp[0][n],sum1+val[n]);
71 }