树形动态规划的框架可以这样写:
Proceduredfs(v);
var i:longint;
Begin
vis[v]:=ture;
for i:=1 to n do
if father[i]=v then// 一般树形动态规划的操作是从叶子节点向根节点推导,所以必须要确定i时v的孩子
begin
if not vis[i]then dfs(i);//如果孩子节点i已经被计算过就不需要再次计算,这里就体现了动态规划的思想
dp[v] fun(dp[i])//这里的分支递归又体现了树的递归操作思想
end;
End;



1039. Anniversary party
Memory Limit: 16 MB
Background
Problem
Input
<L> <K>
0 0
Output
Sample
input | output |
---|---|
7 1 1 1 1 1 1 1 1 3 2 3 6 4 7 4 4 5 3 5 0 0 | 5 |
a[i].in=∑a[i.son].out
a[i].out=∑max{a[i.son].in, a[i.son].out}
in 表示q请 i ,out 表示b不请 i 。
下面说下左儿子右兄弟x新节点的c插入方法:
1。这是一棵二叉树(不大相就是)
2。先帮他认好父亲和兄弟
3。让他爸认识他,并遗弃他的二哥 (为逃过罚款,户口上只能有一个儿子)
#include <iostream>
using namespace std;
struct node {
int boss,son,brother;
int in,out;
void init()
{
boss = brother = son = in = out = 0 ;
}
int max()
{
return in > out ? in : out ;
}
} a[6002];
void treedp(int father){
int son = a[father].son;
while (son) {
treedp(son);
a[father].in+=a[son].out;
a[father].out+=a[son].max();
son=a[son].brother;
}
}
int main(){
int i,l,k,n;
while(cin >> n){
for (i=1; i<=n; i++) {
a[i].init();
cin >> a[i].in;
}
while (cin >>l >>k && l+k) { //边读入建二叉树
a[l].boss = k;
a[l].brother = a[k].son;
a[k].son = l;
}
for (i=1;i<=n;i++)
if (a[i].boss==0) {
treedp(i);
cout << a[i].max() << endl;
break;
}
}
}
也可以不转换二叉树
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=6003;
int a[MAXN];
int father[MAXN];
int f[MAXN][2];
int n;
void TreeDP(int i)
{
f[i][1]=a[i];
for(int j=1;j<=n;j++)
{
if(father[j]==i)
{
TreeDP(j);
f[i][0]+=max(f[j][0],f[j][1]);
f[i][1]+=f[j][0];
}
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",a+i);
}
int x,s;
memset(father,0,sizeof(father));
memset(f,0,sizeof(f));
while(scanf("%d %d",&x,&s),x&&s)
{
father[x]=s;
}
for(int i=1;i<=n;i++)
{
if(father[i]==0)
{
TreeDP(i);
printf("%d\n",max(f[i][0],f[i][1]));
break;
}
}
return 0;
}
http://hi.baidu.com/keynoheart/blog/item/7c820ed0f9a2423e06088b64.html
http://www.cppblog.com/cxiaojia/archive/2011/12/04/ural1039.html