BZOJ4813 CQOI2017小Q的棋盘（树形dp）

　　设f[i][j]为由i号点开始在子树内走j步最多能经过多少格点，g[i][j]为由i号点开始在子树内走j步且回到i最多能经过多少格点，转移显然。

#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<''||c>'')) c=getchar();return c;}

int gcd(int n,int m){return m==?n:gcd(m,n%m);}

int read()

{

    int x=,f=;char c=getchar();

    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}

    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();

    return x*f;

}

int n,m,p[N],f[N][N],g[N][N],t;

struct data{int to,nxt;

}edge[N<<];

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][]=g[k][]=;for (int i=;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>=;x--)

            for (int y=;y+<=x;y++)

            f[k][x]=max(f[k][x],max(g[k][x-y-]+f[edge[i].to][y],(x-y->=?f[k][x-y-]+g[edge[i].to][y]:-n)));

        for (int x=m;x>=;x--)

            for (int y=;y+<=x;y++)

            g[k][x]=max(g[k][x],g[k][x-y-]+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=;i<n;i++)

    {

        int x=read()+,y=read()+;

        addedge(x,y),addedge(y,x);

    }

    dfs(,);

    for (int i=;i<m;i++) f[][m]=max(f[][m],f[][i]);

    cout<<f[][m];

    return ;

}