TreeDp POJ2342 Anniversary party

大学庆典嘉宾名单优化

最新推荐文章于 2019-08-08 13:46:36 发布

原创最新推荐文章于 2019-08-08 13:46:36 发布 · 322 阅读

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为庆祝某大学80周年纪念活动，需要从员工中选出总亲和力评分最高的嘉宾组合参加聚会，但不能同时邀请员工及其直接上司。通过构建员工间的层级关系树，并使用记忆化搜索算法来解决此问题。

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

Sample Output

Source

Ural State University Internal Contest October'2000 Students Session

题解：

树读入后。

可以记忆化搜索。

公式：

treedp[i][0]表不取i的最大rating，treedp[i][1]表取。

所以

1、treedp[i][1]=本身的RATING+treedp[j][0](j是i的儿子）

2. treedp[i][0]+=max(dp(j,0),dp(j,1));
最后取大值即可。

代码：

//Tree DP
#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
struct node{
int son[1001];
int sonnum;
int rating;
}a[6001];
int treedp[6001][2];
bool pg[6001];
int n;
void init()
{
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i].rating;
int x,y;
while (scanf("%d%d",&x,&y)==2&&x*y!=0)
{
a[y].son[++a[y].sonnum]=x;
}
}
int dp(int x,int situation)
{
if (a[x].sonnum==0&&situation) return a[x].rating;
if (treedp[x][situation]) return treedp[x][situation];
if(situation==1)
{
for (int i=1;i<=a[x].sonnum;i++)
{
int j=a[x].son[i];
treedp[x][situation]+=dp(j,0);
}
treedp[x][situation]+=a[x].rating;;
}
if(situation==0)
{
for (int i=1;i<=a[x].sonnum;i++)
{
int j=a[x].son[i];
treedp[x][situation]+=max(dp(j,0),dp(j,1));
}
}
return treedp[x][situation];
}
int main()
{
init();
int ans=a[1].rating;
for(int i=1;i<=n;i++)
{
ans=max(ans,dp(i,1));
ans=max(ans,dp(i,0));
}
cout<<ans;
}