hdu1520 第一道树形DP，激动哇咔咔！

A - 树形dp

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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 
题目大意：给你一 棵关系树，让你从中选择若干个人，这些人之间不能有直接的上下级关系，要求最后的到的权值最大
解析见代码：

/*
 hdu1520
 简单树状DP
 解题包括以下几步
 1.关系树的建立（用一个结构体来储存某个节点所包含的信息。）
 2.确定状态转移方程 f[i][0]表示不选该节点，f[i][1]表示选择该节点，子树能够得到的最大权值
 j为i的子节点，f[i][1]=1+f[j][0],f[i][0]=max(f[j][0],f[j][1])
 最后答案就是max(f[root][1],f[root][0])
 3.编程实现，记忆化搜索，相当于树的后序遍历（从子到根）
*/
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int maxn = 6000 + 10;
int fa[maxn];
int f[maxn][2];
int indegree[maxn];
bool vis[maxn];
int v[maxn];
int n;
void dfs(int x)
{
    vis[x]=true;
    for(int i=1;i<=n;i++)
    {
        if(!vis[i]&&fa[i]==x)
        {
            dfs(i);
            f[x][0]+=max(f[i][1],f[i][0]);
            f[x][1]+=f[i][0];
        }
    }
}
int main()
{
    int a,b;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
         scanf("%d",&v[i]);
         memset(indegree,0,sizeof(indegree));
         memset(outdegree,0,sizeof(outdegree));
        while(scanf("%d%d",&a,&b))
        {
            if(a==0&&b==0) break;
            fa[a]=b;
            indegree[a]++;
        }
        memset(f,0,sizeof(f));
        memset(vis,false,sizeof(vis));
        for(int i=1;i<=n;i++)//边界初始化
        {
           f[i][1]=v[i];
        }
       // cout<<indegree[5]<<endl;
        for(int i=1;i<=n;i++)
        {
            if(indegree[i]==0)
            {
                dfs(i);
                printf("%d
",max(f[i][0],f[i][1]));
                break;
            }
        }
    }
    return 0;
}