hdu 1250 樹形DP

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Anniversary partyTime Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64uSubmit StatusAppoint description: System Crawler  (2014-07-27)

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

711111111 32 36 47 44 53 50 0  

Sample Output

5  

輸入： 

輸入n個結點，接下去的n行，表示1-n的每個結點分別具有的活躍值，在接下來去的n-1行，輸入a,b，表示b是a的上司

輸出：

由於直接有上司和下屬關係的兩個人不能同時參加party， 求出能讓party活躍值最大的方案（求出最大的活躍值即可）.

#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;}