POJ 2342 树形DP

Anniversary party

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8184		Accepted: 4685

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 N – 1 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

题意：

有n个点形成的树，每个点有一个价值，要选最多的价值，其中儿子节点和直接父节点不能同时选，最多价值是多少。

代码：

//考虑每个点选或不选两种状态，递归得到答案。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int n,dp[6010][2],head[6010],tol;
struct node{
    int u,v,next;
}nodes[15000];
void Add(int x,int y){
    nodes[tol].u=x;
    nodes[tol].v=y;
    nodes[tol].next=head[x];
    head[x]=tol++;
}
void dfs(int rt,int fa){
    int max1=0,max2=0;
    for(int i=head[rt];i!=-1;i=nodes[i].next){
        int son=nodes[i].v;
        if(son==fa) continue;
        dfs(son,rt);
        max1+=max(dp[son][1],dp[son][0]);
        max2+=dp[son][0];
    }
    dp[rt][0]=max1;
    dp[rt][1]+=max2;
}
int main()
{
    int x,y;
    memset(dp,0,sizeof(dp));
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&x);
        dp[i][1]=x;
    }
    tol=0;
    memset(head,-1,sizeof(head));
    while(scanf("%d%d",&x,&y)&&x&&y){
        Add(x,y);
        Add(y,x);
    }   
    dfs(1,0);
    printf("%d
",max(dp[1][0],dp[1][1]));
    return 0;
}