hdu 1520 Anniversary party（树dp）

Anniversary party

[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3354 Accepted Submission(s): 1528

[/b]

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

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题意:某大学有N个职员，编号为1~N。他们有从属关系，也就是说他们的关系就像一棵以校长为根的树，父结点就是子结点的直接上司。每个职员有一个快乐指数。现在有个周年庆宴会，要求与会职员的快乐指数最大。但是，没有职员愿和直接上司一起与会。

题解：树dp

#include<stdio.h>
int fath[6008],val[6008],n;
int all,head[6008],dp[6008][2];
struct edge{
int next,data;
}p[6888];
int MAX(int a,int b){ return a>b?a:b; }
void add(int x,int y)
{
p[all].next=head[x];
p[all].data=y;
head[x]=all++;
}
void dfs(int x)
{
int temp,i;

dp[x][0]=0;  dp[x][1]=val[x];
if(head[x]==-1) return;
for(i=head[x];i!=-1;i=p[i].next)
{
temp=p[i].data;
dfs(temp);
dp[x][0]+=MAX(dp[temp][0],dp[temp][1]);
dp[x][1]+=dp[temp][0];
}
}
int main()
{
int i,x,y;

while(scanf("%d",&n)>0)
{
for(i=1;i<=n;i++)
{
scanf("%d",val+i);
head[i]=fath[i]=-1;
}
for(all=i=0;i<n-1;i++)
{
scanf("%d%d",&x,&y);
fath[x]=y;  add(y,x);
}
scanf("%d%d",&x,&y);
for(i=1;i<=n;i++)
{
if(fath[i]!=-1) continue;
dfs(i);  break;
}
printf("%d\n",MAX(dp[i][0],dp[i][1]));
}

return 0;
}