POJ 2486 经典树形dp

Apple Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6654		Accepted: 2197

Description
Wshxzt is a lovely girl. She likes apple very much. One day HX takes her to an apple tree. There are N nodes in the tree. Each node has an amount of apples. Wshxzt starts her happy trip at one node. She can eat up all the apples
in the nodes she reaches. HX is a kind guy. He knows that eating too many can make the lovely girl become fat. So he doesn’t allow Wshxzt to go more than K steps in the tree. It costs one step when she goes from one node to another adjacent node. Wshxzt likes
apple very much. So she wants to eat as many as she can. Can you tell how many apples she can eat in at most K steps.
Input
There are several test cases in the input

Each test case contains three parts.

The first part is two numbers N K, whose meanings we have talked about just now. We denote the nodes by 1 2 ... N. Since it is a tree, each node can reach any other in only one route. (1<=N<=100, 0<=K<=200)

The second part contains N integers (All integers are nonnegative and not bigger than 1000). The ith number is the amount of apples in Node i.

The third part contains N-1 line. There are two numbers A,B in each line, meaning that Node A and Node B are adjacent.

Input will be ended by the end of file.

Note: Wshxzt starts at Node 1.
Output
For each test case, output the maximal numbers of apples Wshxzt can eat at a line.
Sample Input

2 1
0 11
1 2
3 2
0 1 2
1 2
1 3

Sample Output

11
2

题意：一个n个节点的树，每个节点都有一个权值，从1出发，走k步，所能得到的最大权值，每个节点的权值只能算一次。

经典树形dp，理解了好久，终于看懂了。

代码：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-4 14:10:17
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int head[110],tol,dp[110][310][2],weight[110],m;
struct node{
int next,to;
}edge[300];
void add(int u,int v){
edge[tol].to=v;
edge[tol].next=head[u];
head[u]=tol++;
}
void dfs(int u,int fa){
for(int i=0;i<=m;i++)
dp[u][i][0]=dp[u][i][1]=weight[u];
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].to;
if(v==fa)continue;
dfs(v,u);
for(int j=m;j>=0;j--)
for(int k=0;k<=j;k++){
dp[u][j+2][0]=max(dp[u][j+2][0],dp[u][j-k][0]+dp[v][k][0]);//u返回，v返回
dp[u][j+2][1]=max(dp[u][j+2][1],dp[u][j-k][1]+dp[v][k][0]);//从u出发到v，v返回经过u访问u的其他子树不返回。
dp[u][j+1][1]=max(dp[u][j+1][1],dp[u][j-k][0]+dp[v][k][1]);//从u的其他子树返回u，再访问u的子树v并不返回。
}
}
}
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int i,j,k,n;
while(~scanf("%d%d",&n,&m)){
memset(head,-1,sizeof(head));tol=0;
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
scanf("%d",&weight[i]);
for(i=1;i<n;i++){
scanf("%d%d",&j,&k);
add(j,k);
add(k,j);
}
dfs(1,-1);
printf("%d\n",dp[1][m][1]);
}
return 0;
}