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HDU 1520 Anniversary party (tree dp)

2022-08-10 11:09:00 【51CTO】

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

Sample Output

题意

输入 n nodes and their active values,然后输入 a b ,代表 b 是 a 的上司,Supervisors and subordinates who are directly related cannot participate at the same timePatty,求Patty的最大活跃值.

思路

You can first build a tree according to the given relationship.

Each node has two states,To participate or not to participate, 0 Representatives do not participate, 1 代表参加.

我们定义 dp[i][0] 为以 i is a subtree of the root and i The maximum active value you can get by not participating, dp[i][1] 为 i Participate in the maximum active value you can get.

状态转移方程：

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

dp[i][1]+=dp[j][0];

其中 j 是 i 的所有下属.

AC 代码

       
       #include<iostream>
       
       #include<algorithm>
       
       #include<stdio.h>
       
       #include<string.h>
       
       #include<math.h>
       
       #include<iostream>
       
       using 
       
       namespace 
       
       std;
       
       #include<queue>
       
       #include<stack>
       
       int 
       
       N;
       
       int 
       
       GS[
       
       6100][
       
       2];
       
       int 
       
       a[
       
       6100];
       
       vector
       
       <
       
       int
       
       >
       
       G[
       
       6100];
       
       bool 
       
       vis[
       
       6100];
       
       void 
       
       dfs(
       
       int 
       
       x)
       
{
       
       int 
       
       len
       
       =
       
       G[
       
       x].
       
       size();
       
       GS[
       
       x][
       
       0]
       
       =
       
       0;
       
       GS[
       
       x][
       
       1]
       
       =
       
       a[
       
       x];
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <
       
       len; 
       
       i
       
       ++)
       
    {
       
       dfs(
       
       G[
       
       x][
       
       i]);
       
       GS[
       
       x][
       
       0]
       
       +=
       
       max(
       
       GS[
       
       G[
       
       x][
       
       i]][
       
       1],
       
       GS[
       
       G[
       
       x][
       
       i]][
       
       0]);
       
       GS[
       
       x][
       
       1]
       
       +=
       
       GS[
       
       G[
       
       x][
       
       i]][
       
       0];
       
    }
       
}
       
       int 
       
       main()
       
{
       
       while(
       
       ~scanf(
       
       "%d",
       
       &
       
       N)
       
       &&
       
       N)
       
    {
       
       int 
       
       aa,
       
       bb;
       
       for(
       
       int 
       
       i
       
       =
       
       1; 
       
       i
       
       <=
       
       N; 
       
       i
       
       ++)
       
        {
       
       scanf(
       
       "%d",
       
       a
       
       +
       
       i);
       
       G[
       
       i].
       
       clear();
       
        }
       
       memset(
       
       vis,
       
       false,
       
       sizeof(
       
       vis));
       
       while(
       
       ~scanf(
       
       "%d%d",
       
       &
       
       aa,
       
       &
       
       bb)
       
       &&(
       
       aa
       
       ||
       
       bb))
       
        {
       
       G[
       
       bb].
       
       push_back(
       
       aa);
       
       vis[
       
       aa]
       
       =
       
       true;
       
        }
       
       int 
       
       ans
       
       =
       
       0;
       
       for(
       
       int 
       
       i
       
       =
       
       1; 
       
       i
       
       <=
       
       N; 
       
       i
       
       ++)
       
        {
       
       if(
       
       !
       
       vis[
       
       i])
       
            {
       
       dfs(
       
       i);
       
       ans
       
       +=
       
       max(
       
       GS[
       
       i][
       
       0],
       
       GS[
       
       i][
       
       1]);
       
            }
       
        }
       
       printf(
       
       "%d\n",
       
       ans);
       
    }
       
       return 
       
       0;
       
}
      
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