ural 1039 树dp

http://acm.timus.ru/problem.aspx?space=1&num=1039

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

树上的树dp例题，f[i][0]表示以i为根不包含i可获得的最大价值，f[i][1]表示以i为根包含i在内可获得的最大价值。有f[i][0]+=MAX{f[son[i]][0],f[son[i]][1] } ,f[i][1]+=f[son[i]][0];

因为一旦包含i了显然不能包含i的儿子，所以不能加上f[son[i]][1],反之取二者中较大的就好了。

因为如何保存他们之间的关系想了半天，想用vector来着，看书上写的很简便，利用数组做邻接表处理。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define LL long long
 #define inf 0x3f3f3f3f
 int c[],fa[],son[],bro[];
 int dp[][];
 void solve(int id)
 {
     dp[id][]=;
     dp[id][]=c[id];
     for(int i=son[id];i!=-;i=bro[i])
     {
         solve(i);
         dp[id][]+=max(dp[i][],dp[i][]);
         dp[id][]+=dp[i][];
     }
 }
 int main()
 {
     int N,i,j,k;
     while(cin>>N){
         int a,b,root;
         memset(fa,-,sizeof(fa));
         memset(son,-,sizeof(son));
         memset(bro,-,sizeof(bro));
         for(i=;i<=N;++i) scanf("%d",c+i);
         while(scanf("%d%d",&a,&b)&&(a||b)){
             fa[a]=b;
             bro[a]=son[b];
             son[b]=a;
         }
         for(i=;i<=N;++i){
             dp[i][]=,dp[i][]=c[i];
             if(fa[i]==-) root=i;
         }
         solve(root);
         printf("%d\n",max(dp[root][],dp[root][]));
     }
     return ;
 }