E. Tree

狀態表示：$f_{u,j}$表示以$u$節點的子樹，$u$所在連通塊大小為$j$時，并且沒有算上$u$連通塊的貢獻的最大值
狀態計算：
對于一棵子樹$v$來說，顯然可以有兩種情況

• $u$節點與$v$節點不連通：$f_{u,j}=f_{u,j}\times \max [f_{v,1\to s{z}_v}\times (1\to s{z}_v)]$
• $u$節點與$v$節點連通：$f_{u,j+k}=f_{u,j}\times f_{v,k}$

而答案就是$\max [f_{1,1\to s{z}_1}\times (1\to s{z}_1)]$（1是根節點，乘的這部分$1\to s{z}_1$是該節點所在連通塊的貢獻）

---

你以為完了???不不不，還要寫個高精度！！！

#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; }

一下午搞了個這個題，順便整理了一下高精度摸吧yep！
要加油哦~

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