HDU - 1520 Anniversary party 树形dp

Anniversary party

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10129    Accepted Submission(s): 4289

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

第一道树形dp

dp【i】【0】是第i个人不来时的“”rating”最大值

dp【i】【1】是第i个人来时的“rating”最大值

所以dp【i】【0】=∑max（dp【j】【0】，dp【j】【1】）   ，其中j是i的直系下属，第一次写的时候忘了注意要取这两个里的最大值，因为i不来时，j可以来也可以不来

dp【i】【1】=∑dp【j】【0】  （j是i的直系下属）

用vector存每个人的直系下属

类似后序遍历这颗树即可

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
typedef long long ll;
using namespace std;
int n;
vector <int> w[10000];
int l,k,dp[10000][2],book[10000],fa[10000];
int tree_dp(int x)
{
	int t;
	book[x]=1;
	for(int i=0;i<w[x].size();i++)
	{
		t=w[x].at(i);
		if(fa[t]==x&&!book[t])
		{
		//	printf("%d\n",t);
			tree_dp(t);
			dp[x][0]+=max(dp[t][1],dp[t][0]);
			dp[x][1]+=dp[t][0];
		}
	}
}
int main()
{
	int i;
	while(~scanf("%d",&n))
	{
		memset(book,0,sizeof(book));
		for(i=1;i<=n;i++)
		{
			scanf("%d",&dp[i][1]);
			dp[i][0]=0;
			fa[i]=i;	//为了找根节点，开了一个数组存他的直系父节点，应该也有别的简单的求根节点的方法
			w[i].clear();
		}
		while(scanf("%d %d",&l,&k),l,k)
		{
			w[k].push_back(l);
			fa[l]=k;
		}
		for(i=1;i<=n;i++)
		{
			if(fa[i]==i)
			{
				tree_dp(i);	
				printf("%d\n",max(dp[i][0],dp[i][1]));
			//	break;	
			}
		} 
	}

 }