POJ 2342

Anniversary party

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

Source

Ural State University Internal Contest October‘2000 Students Session

题意：

要举行一个party，一个人和他的直属上司不能同时来，每个人都有一个欢乐值，求最大的欢乐值。 转化为树，即父亲与儿子不能同时存在，典型的树状DP。

确定状态：dp[i][0]表示 i结点不来时，i及其子树最大的欢乐值

dp[i][1]表示 i节点来时，i及其子树最大的欢乐值

状态转移方程：

x是i的父节点

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

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

注意初始化 dp[i][0] = 0; dp[i][1] = 1；

结果： ans = max（dp[root][0], dp[root][1]）;

实现：    保存父节点，dfs（）实现记忆化搜索

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxn 6000+5
int n,rat[maxn],fa[maxn],dp[maxn][2],vis[maxn];
void dfs(int x )
{
  vis[x] = 1;
  for(int i = 1;i <= n;i ++)
  {
    if(!vis[i] && fa[i] == x)
    {
      dfs(i);
      dp[x][0] += max(dp[i][0], dp[i][1]);
      dp[x][1] += dp[i][0] ;
    }
  }
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
     int a,b,root;
     memset(vis, 0, sizeof(vis));
     memset(fa, 0, sizeof(fa));
     for(int i = 1;i <= n;i ++)
     scanf("%d", &rat[i]);
     for(int i = 1;i <= n-1;i ++)
     {
       scanf("%d%d", &a, &b);
       fa[a] = b;
     }
     scanf("%d%d", &a, &b);
     for(int i = 1;i <= n;i ++)
     {
       if(fa[i] == 0) root = i;
       dp[i][0] = 0;
       dp[i][1] = rat[i];
     }
     dfs(root);
     printf("%d\n",max(dp[root][0],dp[root][1]));
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。