URAL 1039 Anniversary Party 树形DP 水题

1039. Anniversary Party

Time limit: 0.5 second
Memory limit: 8 MB

Background

The president of the Ural State University is going to make an 80'th Anniversary party. The university has a hierarchical structure of employees; that is, the supervisor relation forms a tree rooted at the president. Employees are numbered by integer numbers in a range from 1 to N, The personnel office has ranked each employee with a conviviality rating. In order to make the party fun for all attendees, the president does not want both an employee and his or her immediate supervisor to attend.

Problem

Your task is to make up a guest list with the maximal conviviality rating of the guests.

Input

The first line of the input contains a number N. 1 ≤ N ≤ 6000. 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 the supervisor relation tree goes. Each line of the tree specification has the form

<L> <K>

which means that the K-th employee is an immediate supervisor of L-th employee. Input is ended with the line

0 0

Output

The output should contain the maximal total rating of the guests.

Sample

input	output
7 1 1 1 1 1 1 1 1 3 2 3 6 4 7 4 4 5 3 5 0 0	5

 

 

题意：一个公司有n个人，根据职位刚好组成一个树。

每一个人都有一个有趣值。

现在要从这n个人中邀请部分人去参加party，使得这个party的有趣值为最大。

但是有个要求：一个人和他的直属上司不能被同时邀请(不然就无趣了)

 

注意：根节点不固定，要根据数据确定。

　　   根节点可能不会被邀请。

 

 

树形DP水题。

dp[i][0]表示i不去的话以i为根的子树的最大的有趣值。

dp[i][1]表示i去的话以i为根的子树的最大的有趣值。

 

树形DP：

1.建树(链式前向星)

2.可能需要DFS预处理

3.递归求解dp

 

 

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<algorithm>

 4 

 5 using namespace std;

 6 

 7 const int maxn=6010;

 8 

 9 struct Edge

10 {

11     int to,next;

12 }edge[maxn];

13 int head[maxn];

14 int tot;

15 int dp[maxn][2];

16 int w[maxn];

17 int in[maxn];

18 int rt;

19 

20 void init()

21 {

22     memset(head,-1,sizeof(head));

23     tot=1;

24     memset(dp,0,sizeof(dp));

25     memset(in,0,sizeof(in));

26 }

27 

28 void addedge(int u,int v)

29 {

30     edge[tot].to=v;

31     edge[tot].next=head[u];

32     head[u]=tot++;

33 }

34 

35 void tree_dp(int u,int fa)

36 {

37     dp[u][1]=w[u];

38     for(int i=head[u];~i;i=edge[i].next)

39     {

40         int v=edge[i].to;

41         if(v==fa)

42             continue;

43         tree_dp(v,u);

44         dp[u][1]+=dp[v][0];

45         dp[u][0]+=max(dp[v][0],dp[v][1]);

46     }

47 }

48 

49 int main()

50 {

51     int n;

52     while(scanf("%d",&n)!=EOF)

53     {

54         init();

55         scanf("%d",&w[1]);

56         if(n==0&&w[1]==0)

57             break;

58         for(int i=2;i<=n;i++)

59             scanf("%d",&w[i]);

60         for(int i=1;i<n;i++)

61         {

62             int u,v;

63             scanf("%d%d",&v,&u);

64             addedge(u,v);

65             in[v]++;

66         }

67         for(int i=1;i<=n;i++)

68             if(in[i]==0)

69             {

70                 rt=i;

71                 break;

72             }

73 

74         tree_dp(rt,-1);

75 

76         int ans=max(dp[rt][0],dp[rt][1]);

77         printf("%d\n",ans);

78     }

79     return 0;

80 }

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