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);//-1，因为cnt初始化为了0
    mem(val,0);
    mem(dp,0);
    mem(num,0);
    cnt=0;
}
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!=-1;i=e[i].nex){
        int v=e[i].to;
        if(val[v]==0){//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]=1;
    for(int i=head[u];i!=-1;i=e[i].nex){
        int v=e[i].to;
        if(v==fa) continue;
        num[u]+=dfs(v,u);
    }
    dp[u][1]=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!=-1;i=e[i].nex){
        int v=e[i].to;
        if(v==fa) continue;
        for(int j=num[u];j>=1;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=1;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=0;i<n-1;i++){
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);//无向图
        }
        val[1]=inf;//不是0就可以
        getval(1);
        val[1]=0;
        dfs(1,0);
        printf("%d
",dp[1][q+1]);//q条边对应q+1个点
    }
    return 0;
}