poj 2486 Apple Tree (树形dp)

Apple Tree

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6674 Accepted: 2208

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 0 111 23 20 1 21 21 3

Sample Output

112

Source

POJ Contest,Author:magicpig@ZSU

题意：

给出一颗苹果树，每个节点有一定数量的苹果，你开始在1号节点，能走m步，问你最多能得到几个苹果。

思路：

树形dp，从每个点走后可以回来或者不回来，可以走k步，构造状态

dp[2][j][k]   01-表示是否回来  j-在哪个点  k-能走几步  dp-能得到的最大苹果数

于是可以分为三种情况：

1.从一个点出发最终回到该点。

2.从一个点出发最终不回该点，走到刚访问了的子树。

3.从一个点出发最终不回该点，走到以前访问过的子树。

转移方程见代码

那么每种状态都能合理转移了，注意循环的顺序。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <map>#include <stack>#include <vector>#include <set>#include <queue>#pragma comment (linker,"/STACK:102400000,102400000")#define maxn 205#define MAXN 400005#define mod 1000000009#define INF 0x3f3f3f3f#define pi acos(-1.0)#define eps 1e-6typedef long long ll;using namespace std;int n,m,k,ans,cnt,tot,flag;int pp[maxn],num[maxn],dp[2][maxn][maxn];struct Node{    int v,w,next;} edge[maxn];void addedge(int u,int v,int w){    cnt++;    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=pp[u];    pp[u]=cnt;}void dfs(int u,int pre){    int i,j,t,v;    for(j=0; j<=m; j++)    {        dp[0][u][j]=dp[1][u][j]=num[u];    }    for(i=pp[u]; i; i=edge[i].next)    {        v=edge[i].v;        if(v!=pre)        {            dfs(v,u);            for(j=m; j>=0; j--)            {                for(k=0; k<=j; k++)                {                    dp[0][u][j+1]=max(dp[0][u][j+1],dp[0][v][k]+dp[1][u][j-k]);                    dp[0][u][j+2]=max(dp[0][u][j+2],dp[1][v][k]+dp[0][u][j-k]);                    dp[1][u][j+2]=max(dp[1][u][j+2],dp[1][v][k]+dp[1][u][j-k]);                }            }        }    }}void solve(){    int i,j,t;    dfs(1,0);    ans=0;    for(i=0; i<=m; i++)    {        ans=max(ans,dp[0][1][i]);        ans=max(ans,dp[1][1][i]);    }}int main(){    int i,j,t;    while(~scanf("%d%d",&n,&m))    {        for(i=1; i<=n; i++)        {            scanf("%d",&num[i]);        }        int u,v;        cnt=0;        memset(pp,0,sizeof(pp));        for(i=1; i<n; i++)        {            scanf("%d%d",&u,&v);            addedge(u,v,1);            addedge(v,u,1);        }        solve();        printf("%d\n",ans);    }    return 0;}/*2 10 111 23 20 1 21 21 3*/