Anniversary party（树形dp）

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5142    Accepted Submission(s): 2367

Problem 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 T 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

 

Recommend

linle   |   We have carefully selected several similar problems for you:   1561  1011  2196  1494  2242 

链接：点击打开链接

// 题意：一棵树每个节点有一个权值，保证父亲节点与孩子节点不能同时出现，求出现的节点最大权值。
// 基础树形dp。
#include <iostream>
using namespace std ;


struct Tree                 //用孩子兄弟表示法存储
{
    int father ;
    int child ;
    int brother ;
    int with_max;    //表示选择该节点的子树所得的最大值
    int without_max;  //表示不选该节点的子树所得的最大值
    int MAX()       //现在再看这些写法，感觉很熟悉了。
    {
        return with_max>=without_max?with_max:without_max;
    }
    void init()
    {
        father=child=brother=without_max=0 ;
    }
}tree[6001] ;


void dfs(int id)
{
    int child;
    child=tree[id].child;
    while(child)
    {
        dfs(child);
        tree[id].with_max+=tree[child].without_max;  //dp---有父亲一定不能有儿子
        tree[id].without_max+=tree[child].MAX();  //这里的tree[child].MAX()不能写成tree[child].with_max，因为没父亲儿子可有可无;
        child=tree[child].brother;
    }
}


int main()
{
    int n,i;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&tree[i].with_max);
            tree[i].init();
        }
        int a,b;
        while(scanf("%d%d",&a,&b),a||b)  //构造孩子兄弟树----很巧妙，注意方法，
        {                                //与书上讲的孩子兄弟树的存储方法不同不同
            tree[a].father=b;
            tree[a].brother=tree[b].child;
            tree[b].child=a; 
        }
        for(i=1;i<=n;i++)
            if(!tree[i].father)  //题目中说的是一个学校的组织，应该不会出现单独的部门吧，
            {                        //所以只会出现一棵树，即只有一根
                dfs(i) ;
                printf("%d\n",tree[i].MAX());
                break ;
            }
    }
    return 0 ;
}

code 2：

题意：

某公司要举办一次晚会，但是为了使得晚会的气氛更加活跃，每个参加晚会的人都不希望在晚会中见到他的直接上司，现在已知每个人的活跃指数和上司关系（当然不可能存在环），求邀请哪些人（多少人）来能使得晚会的总活跃指数最大。

思路：

任何一个点的取舍可以看作一种决策，那么状态就是在某个点取的时候或者不取的时候，以他为根的子树能有的最大活跃总值。分别可以用f[i,1]和f[i,0]表示第i个人来和不来。

当i来的时候，dp[i][1] += dp[j][0];//j为i的下属

当i不来的时候，dp[i][0] +=max(dp[j][1],dp[j][0]);//j为i的下属

以下代码参考：http://hi.baidu.com/saintlleo/blog/item/0606b3feb7026ad3b48f3111.html

//AC CODE:

 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<vector>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<string>

 using namespace std;

 #define maxn 6005

 int n;
 int dp[maxn][2],father[maxn];//dp[i][0]0表示不去，dp[i][1]1表示去了
 bool visited[maxn];

 void tree_dp(int node)
 {
     int i;
     visited[node] = 1;
     for(i=1; i<=n; i++)
     {
         if(!visited[i]&&father[i] == node)//i为下属
         {
             tree_dp(i);//递归调用孩子结点，从叶子结点开始dp
             //关键
             dp[node][1] += dp[i][0];//上司来,下属不来
             dp[node][0] +=max(dp[i][1],dp[i][0]);//上司不来，下属来、不来
         }
     }
 }

 int main()
 {
     int i;
     int f,c,root;
     while(scanf("%d",&n)!=EOF)
     {
         memset(dp,0,sizeof(dp));
         memset(father,0,sizeof(father));
         memset(visited,0,sizeof(visited));
         for(i=1; i<=n; i++)
         {
             scanf("%d",&dp[i][1]);
         }
         root = 0;//记录父结点
         bool beg = 1;
         while (scanf("%d %d",&c,&f),c||f)
         {
             father[c] = f;
             if( root == c || beg )
             {
                 root = f;
             }
         }
         while(father[root])//查找父结点
             root=father[root];
         tree_dp(root);
         int imax=max(dp[root][0],dp[root][1]);
         printf("%d\n",imax);
     }
     return 0;

 }