poj2342 Anniversary party 简单树形dp

http://poj.org/problem?id=2342

Anniversary party

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：每个人都有个价值，然后每个人不和他的直系上属下属一起，问这样产生的最大价值多少。。

题解：入门级的tree dp,dp[i][0]表示以i为根的子树不选i时产生的最大价值，dp[i][1]表示选i产生的。转移是:

dp[i][0]=sum(max(dp[j][0],dp[j][1]))

dp[i][1]+=sum(dp[j][0])     (i是j的父亲)

边界是dp[i][0]=0,dp[i][1]=value[i].

结果就是max(dp[root][0],dp[root][1])

加的是双向边，这样任选一个作为最开始的root都可以

代码:

/**
 * @author neko01
 */
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf("%d\n",a)
typedef pair<int,int> pp;
const double eps=1e-9;
const double pi=acos(-1.0);
const int INF=0x3f3f3f3f;
const LL inf=(((LL)1)<<61)+5;
const int N=6006;
struct node{
    int to,next;
}e[N*2];
int head[N];
int tot;
int dp[N][2];
bool vis[N];
void init()
{
    tot=0;
    clr1(head);
}
void add(int u,int v)
{
    e[tot].to=v;
    e[tot].next=head[u];
    head[u]=tot++;
}
void dfs(int u)
{
    vis[u]=true;
    for(int i=head[u];i!=-1;i=e[i].next)
    {
        int v=e[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()
{
    int n;
    while(~scanf("%d",&n))
    {
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&dp[i][1]);
            dp[i][0]=0;
            vis[i]=false;
        }
        int u,v;
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        scanf("%d%d",&u,&v);
        dfs(1); //加双向边成无根树任意一点都可以
        printf("%d\n",max(dp[1][0],dp[1][1]));
    }
    return 0;
}