HDU 1520 Anniversary party (树形DP)

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2668    Accepted Submission(s): 1192

Problem 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

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

 

 

Source

Ural State University Internal Contest October'2000 Students Session

 

 

Recommend

linle

 

 这道题可以简单的描述为有一些数据之间有父子关系，现在需要选择一些数据出来，使数据总的和最大，但要求数据两两之间不能有直接的父子关系。若设节点i为某一个根，则可以设dp[i][0]表示不包括i的子树的最大和，dp[i][1]表示包括节点i的子树的最大和。则最终所求的结果为：
ans=max{dp[i][0],dp[i][1]}

dp[i][0]=sum(max{dp[j][0],dp[j][1]}) //如果不包括i，则其子树的最大和是不包括其孩子        //或者包括其孩子两种可能下的最大和

 

 

 

 

dp[i][1]=sum(dp[j][0]) //如果包括i，则只能是不包括其孩子的最大和

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int VM=6010;

struct Edge{
    int to,nxt;
}edge[VM<<1];

int cnt,head[VM];
int n,vis[VM],dp[VM][2],happy[VM];

void addedge(int cu,int cv){
    edge[cnt].to=cv;
    edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
}

void DFS(int u){
    if(vis[u])
        return ;
    vis[u]=1;
    dp[u][1]=happy[u];
    for(int i=head[u];i!=-1;i=edge[i].nxt){
        int v=edge[i].to;
        if(!vis[v]){
            DFS(v);
            dp[u][0]+=max(dp[v][0],dp[v][1]);
            dp[u][1]+=dp[v][0];
        }
    }
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d",&n)){
        cnt=0;
        memset(head,-1,sizeof(head));
        for(int i=1;i<=n;i++)
            scanf("%d",&happy[i]);
        int u,v;
        while(~scanf("%d%d",&u,&v)){
            if(u==0 && v==0)
                break;
            addedge(u,v);
            addedge(v,u);
        }
        memset(vis,0,sizeof(vis));
        memset(dp,0,sizeof(dp));
        DFS(1);
        printf("%d\n",max(dp[1][0],dp[1][1]));
    }
    return 0;
}