Apple Tree

Problem 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

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树状dp

分为：

1.一直向前走不回头，用g[x][i]表示为从父节点一直走了i步。

2，走的足够远时，再返回根节点，用f[x][i]表示。

用mark数组标记在议论走过后下一轮不会再重复走父节点。（剪枝）

***************************************************************************************

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<cstdio>
 7 #include<queue>
 8 #include<vector>
 9 #include<stack>
10 using namespace std;
11 vector<int>ve[120];
12 int a[120];
13 int n,m,i,j,k;
14 int f[120][210],g[120][210];
15 int mark[120];
16 int st,en;
17 void dfs(int x)
18  {
19      int i,j;
20      for(i=0;i<ve[x].size();i++)//子节点
21       {
22           int y=ve[x][i];
23           if(mark[y])//跳过
24            continue;
25            mark[y]=1;
26           dfs(y);
27           for(j=m;j>=2;j--)//要返回父节点时，j步时的值
28             f[y][j]=f[y][j-2]+a[y];
29           for(j=m;j>=1;j--)//不返回父节点时，j步时的值
30             g[y][j]=g[y][j-1]+a[y];
31           g[y][0]=f[y][0]=f[y][1]=0;
32           int temp1[210],temp2[210];
33           memset(temp1,0,sizeof(temp1));
34           memset(temp2,0,sizeof(temp2));
35           for(j=0;j<=m;j++)//求最优值
36            for(int k=0;k<=j;k++)
37            {
38                temp1[j]=max(temp1[j],max(g[x][k]+f[y][j-k],g[y][k]+f[x][j-k]));
39                temp2[j]=max(temp2[j],f[x][k]+f[y][j-k]);
40            }
41           for(j=0;j<=m;j++)//赋值
42            {
43                f[x][j]=temp2[j];
44                g[x][j]=temp1[j];
45 
46            }
47       }
48  }
49 int main()
50 {
51     while(scanf("%d%d",&n,&m)!=-1)
52     {
53        for(i=1;i<=n;i++)
54         ve[i].clear();//每次都要清空
55        for(i=1;i<=n;i++)
56          cin>>a[i];
57     for(i=1;i<n;i++)
58      {
59          cin>>st>>en;
60          ve[st].push_back(en);
61          ve[en].push_back(st);
62      }
63      memset(mark,0,sizeof(mark));
64      memset(f,0,sizeof(f));
65      memset(g,0,sizeof(g));
66      mark[1]=1;
67      dfs(1);
68      cout<<g[1][m]+a[1]<<endl;
69     }
70    return 0;
71 }

View Code

转载于:https://www.cnblogs.com/sdau--codeants/p/3273201.html