Anniversary party（树形dp第一步）

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

711111111 32 36 47 44 53 50 0

Sample Output

5

题目大意：

       学校要举办周年纪念日，每个人都有一个欢乐度，但是见到自己的直接上司的时候瞬间就不欢乐了，So，上司去我不去，上司不去我再去，这样才能愉快的玩耍，然后求最大的欢乐度。

解题思路：

       给出一个关系图，把它转化成一个树（小白书上面有，无根树转有根树），然后求出递推方程

                                  dp[i][0] += max(dp[j][0] , dp[j][1]);  i为上司，j为下级                dp[ i ][ 1 ] += dp[ j ][ 0 ];  dp[i][0]表示i不去，dp[i][1]表示i去。

一次遍历求最优解就可以了。

代码如下：

#include<iostream>#include<cstdio>#include<map>#include<math.h>#include<cstring>#include<vector>#include<algorithm>using namespace std;int pre[6010];vector<int> G[6010]; // 动态存储邻接矩阵int dp[6010][2], vis[6010], n;void dfs(int u, int father){    int d = G[u].size();    for(int i = 0; i < d; i++)    {        int v = G[u][i];        if(v != father)        {            dfs(v, pre[v] = u);        }    }}void d(int t){    vis[t] = 1;// 标记已被访问    for(int i = 1; i <= n; i++)    {        if(pre[i] == t && !vis[i])        {            d(i);//当遍历到叶子，求最优解            dp[t][0] += max(dp[i][0], dp[i][1]);            dp[t][1] += dp[i][0];        }    }}int main(){    int i, u, v;    while(scanf("%d",&n) != EOF)    {        memset(dp, 0, sizeof(dp));        for(i = 1; i <= n; i++)            scanf("%d",&dp[i][1]);        while(scanf("%d%d",&u,&v) && (u + v))        {            G[u].push_back(v);            G[v].push_back(u);        }        memset(pre, -1, sizeof(pre));//设置初始都指向-1位父亲结点        dfs(1,-1);        memset(vis, 0, sizeof(vis));        d(1);        printf("%d\n",max(dp[1][0], dp[1][1]));    }    return 0;}