HDU 1250 tree DP

Anniversary Party Time limit:1000 ms Memory limit:32768kb 64bit Io format:% I64d & % i64usubmit statusappoint description: System crawler)

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 shoshould contain the maximal sum of guests ratings.

Sample Input

711111111 32 36 47 44 53 50 0

Sample output

5

Input:

Enter n knots and the next n rows, indicating the active values of each node of 1-n. In the next n-1 line, enter a and B, indicating that B is the boss of.

Output:

Because two people with direct relationship between superiors and subordinates cannot participate in the party at the same time, find the solution that maximizes the activity value of the Party (find the maximum activity value ).

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;const int maxn=6005;vector<int> v[maxn];int value[maxn],in[maxn],dp[maxn][2];inline int max(int a,int b){return a>b?a:b;}void dfs(int id){    dp[id][0]=0;dp[id][1]=value[id];    for(int i=0;i<v[id].size();i++)    {        int u=v[id][i];        dfs(u);        dp[id][0]+=max(dp[u][0],dp[u][1]);        dp[id][1]+=dp[u][0];    }}int main(){    int n,a,b,i,ans;    while(~scanf("%d",&n))    {        for(i=1;i<=n;i++) v[i].clear();        for(i=1;i<=n;i++) scanf("%d",&value[i]);        memset(in,0,sizeof(in));        while(scanf("%d%d",&a,&b),a+b)        {            v[b].push_back(a);in[a]++;        }        memset(dp,0,sizeof(dp));        ans=0;        for(i=1;i<=n;i++)            if(!in[i])            {                dfs(i);                ans+=max(dp[i][0],dp[i][1]);            }        printf("%d\n",ans);    }    return 0;}