POJ：2342-Anniversary party（树形dp入门题目）

传送门：http://poj.org/problem?id=2342

Anniversary party

Time Limit: 1000MS Memory Limit: 65536K

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

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1. 题意很简单就是公司要举办一个party，但是每个人都不想和自己的直属上司一同在party里面，每个人如果参见party有一个欢乐值，要求你计划安排要举办的party欢乐值最大。
2. 其实就是一个树形dp，dp[i][j]代表第i个人是否来（j为1代表来，j为0代表不来），然后就是状态转移方程：dp[root][0] = max(dp[pre_root][1],dp[pre_root][0];
   
   dp[root][1] += dp[pre_root][0];
3. 主要考察的就是一个建树的过程，以及寻找根节点，关于dp的部分还是很简单的。

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#include<stdio.h>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 6010;
int dp[maxn][3],n;
bool vis[maxn];//用来查找根节点
vector<int> ve[maxn];//用来存树

void init()
{
    for(int i=0;i<=n+1;i++)
        ve[i].clear();
    memset(dp,0,sizeof(dp));
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)
        scanf("%d",&dp[i][1]);
    int S,E;
    while(scanf("%d%d",&S,&E) && S+E)
    {
        ve[E].push_back(S);
        vis[S] = true;
    }
}

void dfs(int x)
{
    for(int i=0;i<ve[x].size();i++)
    {
        int temp = ve[x][i];
        dfs(temp);
        //状态转移方程
        dp[x][0] += max(dp[temp][0],dp[temp][1]);
        dp[x][1] += dp[temp][0];
    }
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        init();
        for(int i=1;i<=n;i++)
            if(!vis[i])
            {
                dfs(i);
                printf("%d\n",max(dp[i][0],dp[i][1]));
                break;
            }
    }
    return 0;
}