codeforce23 E. Tree（高精度+树形dp）

E. Tree

状态表示： $f_{u,j}$ 表示以 $u$ 节点的子树， $u$ 所在连通块大小为 $j$ 时，并且没有算上 $u$ 连通块的贡献的最大值
状态计算：
对于一棵子树 $v$ 来说，显然可以有两种情况

$u$ 节点与 $v$ 节点不连通： $fu,j=fu,j×max⁡[fv,1→szv×(1→szv)]f_{u,j}=f_{u,j}×\max[f_{v,1\to sz_{v}}×(1\to sz_v)]$
$u$ 节点与 $v$ 节点连通： $f_{u,j+k}=f_{u,j}×f_{v,k}$

而答案就是 $max⁡[f1,1→sz1×(1→sz1)]\max[f_{1,1\to sz_{1}}×(1\to sz_1)]$ （1是根节点，乘的这部分 $1→sz11\to sz_1$ 是该节点所在连通块的贡献）

你以为完了???不不不，还要写个高精度！！！

#define IO ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(0)
#pragma GCC optimize(2)
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
//==========================================================高精度板子
struct bign
{int d[120], len;bign() { memset(d, 0, sizeof d); len = 1; }bign(int num) { *this = num; }bign(char* num) { *this = num; }bign operator=(const char* num){len = strlen(num);for (int i = 0; i < len; i++) d[i] = num[len - i - 1] - '0';return *this;}bign operator=(int num){char c[1010];sprintf(c, "%d", num);*this = c;return *this;}void clear(){while (len > 1 && !d[len - 1]) len--;}bign operator+(const bign& b){bign c; c.len = 0;for (int i = 0, t = 0; t || i < len || i < b.len; i++){if (i < len) t += d[i];if (i < b.len) t += b.d[i];c.d[c.len++] = t % 10;t /= 10;}return c;}bign operator-(const bign& b){bign c; c.len = 0;for (int i = 0, t = 0; i < len; i++){t += d[i];if (i < b.len) t -= b.d[i];c.d[c.len++] = (t + 10) % 10;if (t >= 0) t = 0;else t = -1;}c.clear();return c;}bign operator*(const bign& b){bign c; c.len = len + b.len;for (int i = 0; i < len; i++) for (int j = 0; j < b.len; j++) c.d[i + j] += d[i] * b.d[j];for (int i = 0; i < c.len - 1; i++) c.d[i + 1] += c.d[i] / 10, c.d[i] %= 10;c.clear();return c;}bool operator < (const bign& b){if (len != b.len) return len < b.len;for (int i = len - 1; i >= 0; i--)if (d[i] != b.d[i]) return d[i] < b.d[i];return false;}bool operator <= (const bign& b){if (len != b.len) return len < b.len;for (int i = len - 1; i >= 0; i--)if (d[i] != b.d[i]) return d[i] < b.d[i];return true;}bool operator > (const bign& b){if (len != b.len) return len > b.len;for (int i = len - 1; i >= 0; i--)if (d[i] != b.d[i]) return d[i] > b.d[i];return false;}bool operator >= (const bign& b){if (len != b.len) return len > b.len;for (int i = len - 1; i >= 0; i--)if (d[i] != b.d[i]) return d[i] > b.d[i];return true;}bign operator+=(const bign& b){*this = *this + b;return *this;}bign operator-=(const bign& b){*this = *this - b;return *this;}bign operator*=(const bign& b){*this = *this * b;return *this;}void print(){for (int i = len - 1; i >= 0; i--) std::cout << d[i];cout << '\n';}string str(){string res = "";for (int i = 0; i < len; i++) res = (char)(d[i] + '0') + res;return res;}
};
istream& operator >>(istream& in, bign& x)
{string s;in >> s;x = s.c_str();return in;
}
ostream& operator <<(ostream& out, bign& x)
{out << x.str();return out;
};
bign max(bign a,bign b)
{return a>b?a:b;
}
//==========================================================
constexpr int N=710;
int h[N],e[2*N],ne[2*N],idx;
void add(int a,int b){e[idx]=b,ne[idx]=h[a],h[a]=idx++;}
int sz[N],n;
bign f[N][N],ans;
void dfs(int u,int fa)
{f[u][1]=1;sz[u]=1;for(int i=h[u];i!=-1;i=ne[i]){int v=e[i];if(v==fa) continue;dfs(v,u);bign now=1;for(int j=1;j<=sz[v];j++) now=max(now,f[v][j]*bign(j));for(int j=sz[u];j>=1;j--){for(int k=sz[v];k>=1;k--)f[u][j+k]=max(f[u][j+k],f[u][j]*f[v][k]);  f[u][j]=f[u][j]*now;    }sz[u]+=sz[v];}for(int j=1;j<=sz[u];j++)ans=max(ans,f[u][j]*bign(j));
}
int main()
{cin>>n;memset(h,-1,sizeof h);for(int i=1;i<n;i++){int u,v;cin>>u>>v;add(u,v),add(v,u);}dfs(1,0);ans.print();return 0;
}