标签：div void play ... com codechef ++ show close
很显然不可能选择砍掉一对有祖先关系的子树。令$f_i$表示$i$子树的答案，如果$i$不被砍，那就是$a_i + \sum\limits_j f_j$；如果$i$被砍，那就是$-x$。取个$max$就好了。
时间复杂度$O(n)$。
#include <bits/stdc++.h> using namespace std; const int N = 1e5 + 5; int tc, n, xx; int a[N]; vector<int> g[N]; long long f[N]; void Dfs(int x, int ft) { f[x] = a[x]; for (int i = 0; i < g[x].size(); ++i) { int v = g[x][i]; if (v == ft) continue; Dfs(v, x); f[x] += f[v]; } f[x] = max(f[x], -(long long)xx); } int main() { scanf("%d", &tc); for (; tc--; ) { scanf("%d%d", &n, &xx); for (int i = 1; i <= n; ++i) { scanf("%d", &a[i]); } for (int i = 1, x, y; i < n; ++i) { scanf("%d%d", &x, &y); g[x].push_back(y); g[y].push_back(x); } Dfs(1, 0); printf("%lld\n", f[1]); // remember to clear up for (int i = 1; i <= n; ++i) { g[i].clear(); } } return 0; }
在模$m$意义下，$a * k(k \in \mathbb{N})$能表示的最大的数就是$m - (a, m)$。容易推导出一个叶子的答案就是$m_i - (m, a_{b_1}, a_{b_2}, ... , a_{b_w})$，其中$b$表示$i$号点的祖先链。
时间复杂度$O(nlogn)$。
#include <bits/stdc++.h> using namespace std; const int N = 1e5 + 5; int tc, n; vector<int> g[N]; long long a[N], m[N], gcd[N]; void Dfs(int x, int ft) { for (int i = 0; i < g[x].size(); ++i) { int v = g[x][i]; if (v == ft) continue; gcd[v] = __gcd(gcd[x], a[v]); Dfs(v, x); } } int main() { scanf("%d", &tc); for (; tc--; ) { scanf("%d", &n); for (int i = 1, x, y; i < n; ++i) { scanf("%d%d", &x, &y); g[x].push_back(y); g[y].push_back(x); } for (int i = 1; i <= n; ++i) { scanf("%lld", &a[i]); } for (int i = 1; i <= n; ++i) { scanf("%lld", &m[i]); } gcd[1] = a[1]; Dfs(1, 0); for (int i = 2; i <= n; ++i) { if (g[i].size() == 1) { long long d = __gcd(gcd[i], m[i]); printf("%lld ", m[i] - d); } } printf("\n"); // remember to clear up for (int i = 1; i <= n; ++i) { g[i].clear(); } } return 0; }
【Code Chef】April Challenge 2019
标签：div void play ... com codechef ++ show close
原文地址：https://www.cnblogs.com/Dance-Of-Faith/p/10712425.html
