設f[i][j]為由i號點開始在子樹內走j步最多能經過多少格點，g[i][j]為由i號點開始在子樹內走j步且回到i最多能經過多少格點，轉移顯然。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 110
char getc(){char c=getchar();while
 
 ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1
 
)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,m,p[N],f[N][N],g[N][N],t;
struct data{int to,nxt;
}edge[N<<1];
void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}
void dfs(int k,int from)
{
    f[k][0]=g[k][0]=1;for (int i=1;i<=m;i++) f[k][i]=g[k][i]=-n;
    for (int
 
 i=p[k];i;i=edge[i].nxt)
    if (edge[i].to!=from)
    {
        dfs(edge[i].to,k);
        for (int x=m;x>=1;x--)
            for (int y=0;y+1<=x;y++)
            f[k][x]=max(f[k][x],max(g[k][x-y-1]+f[edge[i].to][y],(x-y-2>=0?f[k][x-y-2]+g[edge[i].to][y]:-n)));
        for (int x=m;x>=2;x--)
            for (int y=0;y+2<=x;y++)
            g[k][x]=max(g[k][x],g[k][x-y-2]+g[edge[i].to][y]);
    }
    
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4813.in","r",stdin);
    freopen("bzoj4813.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<n;i++)
    {
        int x=read()+1,y=read()+1;
        addedge(x,y),addedge(y,x);
    }
    dfs(1,1);
    for (int i=0;i<m;i++) f[1][m]=max(f[1][m],f[1][i]);
    cout<<f[1][m];
    return 0;
}

BZOJ4813 CQOI2017小Q的棋盤（樹形dp）