HDOJ 题目1520 Anniversary party（树形dp）

Posted 2020-09-23 cxchanpin

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6271    Accepted Submission(s): 2820

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

php?

field=problem&key=Ural+State+University+Internal+Contest+October%272000+Students+Session&source=1&searchmode=source">Ural State University Internal Contest October‘2000 Students Session

 

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题目大意：有n个人去參加party，每一个人都有自己的价值。后边跟很多行a b（0,0结束），表示b是a的上司，不能使他和他的上司同一时候參加，能够获得最大的快乐值，

dp[i][0]表示他不參加的最大快乐值，dp[i][1]表示他參加最大的快乐值

 ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?
a:b)
struct s
{
	int u,v,next;
}edge[3000100];
int a[6060],dp[6060][2],dig[6060],head[6060],vis[6060],cnt;
void add(int u,int v)
{
	edge[cnt].u=u;
	edge[cnt].v=v;
	edge[cnt].next=head[u];
	head[u]=cnt++;
}
void dfs(int u)
{
	vis[u]=1;
	dp[u][1]=a[u];
	int i;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		if(!vis[v])
		{
			dfs(v);
			dp[u][0]+=max(dp[v][0],dp[v][1]);
			dp[u][1]+=dp[v][0];
		}
	}
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int i;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
		}
		int u,v;
		cnt=0;
		memset(head,-1,sizeof(head));
		memset(dp,0,sizeof(dp));
		memset(dig,0,sizeof(dig));
		memset(vis,0,sizeof(vis));
		while(scanf("%d%d",&v,&u),u||v)
		{
			add(u,v);
			dig[v]++;
		}
		int s;
		for(i=1;i<=n;i++)
		{
			if(!dig[i])
			{
				s=i;
			}
		}
		dfs(s);
		printf("%d\n",max(dp[s][0],dp[s][1]));
	}
}