hdu1520 Anniversary party(树形dp)

[align=center]象盐城大数据竞赛！ [/align]

Anniversary party Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 13619    Accepted Submission(s): 5384 [align=left]Problem Description[/align]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.   [align=left]Input[/align]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  [align=left]Output[/align]Output should contain the maximal sum of guests' ratings.   [align=left]Sample Input[/align] 7 1 1 1 1 1 1 1 1 3 2 3 6 4 7 4 4 5 3 5 0 0  [align=left]Sample Output[/align] 5

树形dp入门题
题意：有n个人，每个人有一个欢乐度，下面给出许多父子关系，现在要举行宴会，当父与子同时到场时他们会不开心，求最大的欢乐度
对于每个人，有到与不到两种状态，我们可以用0和1来标记
可得状态方程 
dp[fa][1]+=dp[son][0]  //父亲来
d555
，则该点的权等于所有儿子不来的最大权和
dp[fa][0]+=max(dp[son][1],dp[son][0])  //父亲不来，则该点的权等于max（所有儿子来或不来）权和
首先用并查集找了一下根
然后通过递归，先求出叶节点的值，由叶向根回推即可#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<math.h>
#include<vector>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int N=6010;
const int M=400010;
int f
,dp
[2],a
;
vector<int>e
;
int getf(int x)
{
return x==f[x]?x:f[x]=getf(f[x]);
}
void mix(int x,int y)
{
int tx=getf(x);
int ty=getf(y);
if(tx!=ty)
{
f[tx]=ty;
}
}
void dfs(int x)
{
dp[x][1]+=a[x];
for(int i=0; i<e[x].size(); i++)
{
int v=e[x][i];
dfs(v);
dp[x][0]+=max(dp[v][0],dp[v][1]);
dp[x][1]+=dp[v][0];
}
}
int main()
{
int n,x,y,root;;
while(~scanf("%d",&n))
{
mem(dp,0);
for(int i=1; i<=n; i++)scanf("%d",&a[i]);
for(int i=1; i<=n; i++)
{
e[i].clear(),f[i]=i;
}
while(scanf("%d%d",&x,&y)&&x+y)
{
e[y].push_back(x);
mix(x,y);
}

for(int i=1; i<=n; i++)
{
if(i==getf(i))
{
root=i;
break;
}
}
dfs(root);
printf("%d\n",max(dp[root][0],dp[root][1]));
}
}