【树形dp】Apple Tree

[poj2486]Apple Tree

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10800		Accepted: 3629

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

Source

POJ Contest,Author:magicpig@ZSU

 

题目大意：从N个点的树的根(1)出发，走V步，每个点都有一个权值，问能经过的最大权值和是多少？

试题分析：树形dp一般情况下是有一颗树，询问这棵树的某些最优性信息或者方案数信息。

　　　　　本题有N与V所以至少有两个状态：dp[N][V];

　　　　　考虑dp[i][j]的转移：dp[i][j]=max(dp[i->son][v-1])+a[i]

　　　　　但是这样我们就会发现一个问题：如果是以现在的i为最近公共祖先的两个点呢？是不是没有被包括？

　　　　　所以这样的状态不够，我们设计dp[N][V][2]的状态，01表示回不回i节点

　　　　　所以得出:dp[i][j][0]=max(dp[i][j][0],dp[i->son][t-1][0]+dp[i][j-t][1]+a[i]);//从其它子树转了一圈回来再转到这可子树并不回去了。

　　　　　　　　　 dp[i][j][0]=max(dp[i][j][0],dp[i->son][t-1][1]+dp[i][j-t][0]+a[i])//从这颗子树回到i再到其它子树不会来了

　　　　　　　　　 dp[i][j][1]=max(dp[i][j][1],dp[i->son][t-1][1]+dp[i][j-t][1]+a[i])//走回i

 

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<algorithm>
using namespace std;

inline int read(){
	int x=0,f=1;char c=getchar();
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	return x*f;
}
const int MAXN=100001;
const int INF=999999;
int N,V;
vector<int> vec[301];
int dp[301][301][2];
int a[301];

void dfs(int x,int fa){
	for(int i=0;i<vec[x].size();i++){
		if(vec[x][i]==fa) continue;
		dfs(vec[x][i],x);
		int SON=vec[x][i];
		for(int v=V;v>=1;v--){
			for(int k=1;k<=v;k++){
				dp[x][v][0]=max(dp[x][v][0],dp[x][v-k][1]+dp[SON][k-1][0]);//从别的子树兜了一圈到这可子树，不回去了
				dp[x][v][0]=max(dp[x][v][0],dp[x][v-k][0]+dp[SON][k-2][1]);//从这颗子树兜了一圈到别的子树，不回去了
				dp[x][v][1]=max(dp[x][v][1],dp[x][v-k][1]+dp[SON][k-2][1]);//从别的子树兜了一圈到这可子树然后还回去
			}
		}
	}
}
int ans;

int main(){
	while(scanf("%d%d",&N,&V)!=EOF){
		for(int i=1;i<=300;i++) vec[i].clear();
	    memset(dp,0,sizeof(dp));
		ans=0;
    	for(int i=1;i<=N;i++) a[i]=read();
    	if(N==1){//注意要特判
    		printf("%d\n",a[1]);
    		continue;
		}
	    for(int i=1;i<N;i++){
    		int u=read(),v=read();
    		vec[u].push_back(v);
    		vec[v].push_back(u);
		}
		for(int i=1;i<=N;i++)
	    	for(int j=0;j<=V;j++){
	    		dp[i][j][0]=dp[i][j][1]=a[i];
			}
		dfs(1,0);
	    printf("%d\n",max(dp[1][V][0],dp[1][V][1]));
	}
}