POJ2486 Apple Tree

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

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

若当前走到了结点x，已经走了y步，除了走向子树外，还有另一选择：返回父节点，去父节点的其他子树。

所以比一般的树状DP多加一维状态，记录有没有返回当前结点。0表示要回，1表示不回。

懒得写分析了，复制一份233

dp[root][j][0] = MAX (dp[root][j][0] , dp[root][j-k][0] + dp[son][k-2][0]);//从s出发，要回到s，需要多走两步s-t,t-s,分配给t子树k步，其他子树j-k步，都返回

dp[root][j]][1] = MAX(  dp[root][j][1] , dp[root][j-k][0] + dp[son][k-1][1]) ;//先遍历s的其他子树，回到s，遍历t子树，在当前子树t不返回，多走一步

dp[root][j][1] = MAX (dp[root][j][1] , dp[root][j-k][1] + dp[son][k-2][0]);//不回到s（去s的其他子树），在t子树返回，同样有多出两步

by键盘上的舞者

我实际写的时候0和1与上面说的相反。

 /*by SilverN*/
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 using namespace std;
 const int mxn=;
 struct edge{
     int v;
     int nxt;
 }e[mxn];
 int hd[mxn],mct=;
 void add_edge(int u,int v){
     e[++mct].v=v;e[mct].nxt=hd[u];hd[u]=mct;
     return;
 }
 int f[mxn][mxn][];//第三维0表示不返回根节点，1表示返回根节点
 int w[mxn];
 int n,m;
 void dp(int u,int fa){
     int i,j,k;
     for(i=hd[u];i;i=e[i].nxt){
         int v=e[i].v;
         if(v==fa)continue;
         dp(v,u);
         for(j=m;j;--j){
             for(k=;k<=j;++k){
                 f[u][j][]=max(f[u][j][],f[u][j-k][]+f[v][k-][]);
                 f[u][j][]=max(f[u][j][],f[u][j-k][]+f[v][k-][]);
                 f[u][j][]=max(f[u][j][],f[u][j-k][]+f[v][k-][]);
             }
         }
     }
     return;
 }
 int main(){
     while(scanf("%d%d",&n,&m)!=EOF){
         memset(e,,sizeof e);
         memset(hd,,sizeof hd);
         memset(f,,sizeof );
         mct=;
         int i,j;
         for(i=;i<=n;i++){
             scanf("%d",&w[i]);
             for(j=;j<=m;j++){
                 f[i][j][]=f[i][j][]=w[i];
             }
         }
         int u,v;
         for(i=;i<n;i++){
             scanf("%d%d",&u,&v);
             add_edge(u,v);
             add_edge(v,u);
         }
         dp(,);
         printf("%d\n",f[][m][]);
     }
     return ;
 }