思路
求分数最大值,显然分数规划。接下来就转化成求树上包含根的权值最大联通块。树形DP即可。
代码
#include<bits/stdc++.h>
#define rep(i,st,ed) for(int i=st;i<=ed;++i)
#define per(i,st,ed) for(int i=st;i>=ed;--i)
#define bl(u,i) for(int i=head[u];i;i=e[i].nxt)
#define en puts("")
#define LLM LONG_LONG_MAX
#define LLm LONG_LONG_MIN
#define pii pair<ll,ll>
typedef long long ll;
typedef double db;
using namespace std;
const ll INF=0x3f3f3f3f;
void read() {}
void OP() {}
void op() {}
template <typename T, typename... T2>
inline void read(T &_, T2 &... oth)
{
int __=0;
_=0;
char ch=getchar();
while(!isdigit(ch))
{
if(ch=='-')
__=1;
ch=getchar();
}
while(isdigit(ch))
{
_=_*10+ch-48;
ch=getchar();
}
_=__?-_:_;
read(oth...);
}
template <typename T>
void Out(T _)
{
if(_<0)
{
putchar('-');
_=-_;
}
if(_>=10)
Out(_/10);
putchar(_%10+'0');
}
template <typename T, typename... T2>
inline void OP(T _, T2... oth)
{
Out(_);
putchar('\n');
OP(oth...);
}
template <typename T, typename... T2>
inline void op(T _, T2... oth)
{
Out(_);
putchar(' ');
op(oth...);
}
/*#################################*/
const ll N=2550;
ll k,n;
ll a[N],b[N],siz[N];
db f[N][N],c[N];
vector<ll> e[N];
void dfs(ll u)
{
db now[N],pre[N];
now[siz[u]=1]=c[u];
for(auto v:e[u])
{
dfs(v);
rep(i,1,siz[u])
pre[i]=now[i];
rep(i,2,siz[u]+siz[v])
now[i]=-INF;
rep(i,1,siz[u])
rep(j,0,siz[v])
now[i+j]=max(now[i+j],pre[i]+f[v][j]);
siz[u]+=siz[v];
}
rep(i,1,siz[u])
f[u][i]=now[i];
}
ll check(db x)
{
rep(i,1,n)
c[i]=b[i]-a[i]*x;
dfs(0);
return f[0][k+1]>=0;
}
int main()
{
read(k,n);
ll fa;
rep(i,1,n)
{
read(a[i],b[i],fa);
e[fa].emplace_back(i);
}
db L=0,R=1e4;
rep(i,1,50)
{
db mid=(L+R)/2;
if(check(mid))
L=mid;
else
R=mid;
}
printf("%.3lf",L);
}