树形 dp

// ACM训练联盟周赛     C. Teemo's tree problem

There is an apple tree in Teemo's yard. It contains n nodes and n-1 branches, and the node 1 is always the root of the tree. Today, Teemo's father will go out for work. So Teemo should do his father's job in the family: Cut some branches to make the tree more beautiful. His father's told him that he should cut some branches, finally, the tree should just contains q branches. But when Teemo start to cut, he realizes that there are some apples in the branches( For example, there are 10 apples in the branches which connecting node 1 and node 4). So Teemo not only wants to achieve his father's order, but also wants to preserve apples as much as possible. Can you help him?

 

 

 

 

 

1

2   5

2

 \ / 

3

  3   4

4

   \ /

5

    1

 

 

Input Format

• The first line of the input contains an integer T(1<=T<=10) which means the number of test cases.

• For each test case, The first line of the input contains two integers n,q(3<=n<=100,1<=q<=n-1), giving the number of the node and the number of branches that the tree should preserve.
• In the next n-1 line, each line contains three integers u,v,w(1<=u<=n,1<=v<=n,u!=v,1<=w<=100000), which means there is a branch connecting node u and node v, and there are w apple(s) on it.

Output Format

Print a single integer, which means the maximum possible number of apples can be preserved.

样例输入

1
5 2
1 3 1
1 4 10
2 3 20
3 5 20

样例输出

21

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<vector>
 #define N 209
 using namespace std;
 #define ll long long
 #define mem(a,b)  memset(a,b,sizeof(a))
 #define inf 0x3f3f3f3f
 int t,n,q;
 int u,v,w,head[N],val[N];
 int num[N];//num[i]: 包含i在内及其后面所有点的个数
 int dp[N][N]; //dp[i][j] :以i为“根”，取j个点可获的最大价值
 struct Edge{
     int from,to,nex,w;
 }e[N];
 int cnt;
 void init()
 {
     mem(head,-);//-1，因为cnt初始化为了0
     mem(val,);
     mem(dp,);
     mem(num,);
     cnt=;
 }
 void add(int u,int v,int w){
     e[cnt].from=u;
     e[cnt].to=v;
     e[cnt].w=w;
     e[cnt].nex=head[u];
     head[u]=cnt++;
 }
 void getval(int u){
     for(int i=head[u];i!=-;i=e[i].nex){
         int v=e[i].to;
         if(val[v]==){//找过的不会在找了
         val[v]=e[i].w;//除根节点外把边的值赋给点
         getval(v);
         }
     }
 }
 int  dfs(int u,int fa)
 {
     num[u]=;
     for(int i=head[u];i!=-;i=e[i].nex){
         int v=e[i].to;
         if(v==fa) continue;
         num[u]+=dfs(v,u);
     }
     dp[u][]=val[u];//只有自己，因此val[1]==0 :去掉所有的边才会只有1
     for(int i=head[u];i!=-;i=e[i].nex){
         int v=e[i].to;
         if(v==fa) continue;
         for(int j=num[u];j>=;j--){//逆序，若正序比如 j==1 ,无法更新，后面的也会出现错误
             for(int k=;k<j&&k<=num[v];k++){
                 dp[u][j]=max(dp[u][j],dp[u][j-k]+dp[v][k]);
             }
         }
     }
     return num[u];
 }
 int  main()
 {
     scanf("%d",&t);
     while(t--)
     {
         init();
         scanf("%d%d",&n,&q);
         for(int i=;i<n-;i++){
             scanf("%d%d%d",&u,&v,&w);
             add(u,v,w);
             add(v,u,w);//无向图
         }
         val[]=inf;//不是0就可以
         getval();
         val[]=;
         dfs(,);
         printf("%d\n",dp[][q+]);//q条边对应q+1个点
     }
     return ;
 }