F - Anniversary party

HDU - 1520

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
7
1
1
1
1
1
1
1
1 3
2 3
6 4
7 4
4 5
3 5
0 0
Sample Output
5

題意：一個(gè)公司的領(lǐng)導(dǎo)結(jié)構(gòu)成樹(shù)形結(jié)構(gòu)，每個(gè)人都有一個(gè)領(lǐng)導(dǎo)，現(xiàn)在要開(kāi)一個(gè)party，想邀請(qǐng)互相不是上下級(jí)關(guān)系的人來(lái)參加，每個(gè)人都有一個(gè)參加party的開(kāi)心度，怎么選人使開(kāi)心值最大，輸出最大開(kāi)心值

解法：樹(shù)形dp，對(duì)于每個(gè)結(jié)點(diǎn)有dp[i][0]為此人不參加的最大快樂(lè)值，dp[i][1]為此人參加的最大快樂(lè)值，dp[i][1]+=dp[i-1][1]，dp[i][0]=max(dp[i-1][0],dp[i-1][1])

代碼：

#include<iostream>
#include<vector>
using namespace std;
int dp[6005][2];
bool root[6005];
vector<int> a[6005];
void dfs(int b){
    if(a[b].empty())
        return;
    for(int i=0;i<a[b].size();i++){
        dfs(a[b][i]);
        dp[b][1]+=dp[a[b][i]][0];
        dp[b][0]+=max(dp[a[b][i]][0],dp[a[b][i]][1]);
    }
    return;
}
int main()
{
    int n,x,y,r;
    while(cin>>n){
        for(int i=1;i<=n;i++){
            cin>>dp[i][1];
            dp[i][0]=0;
            root[i]=1;
            a[i].clear();
        }
        while(cin>>x>>y&&x&&y){
            a[y].push_back(x);
            root[x]=0;
        }
        for(int i=1;i<=n;i++){
            if(root[i]==1){
                r=i;
                break;
            }
        }
        dfs(r);
        cout<<max(dp[r][0],dp[r][1])<<endl;
    }
    return 0;
}