Subtree Removal
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。
#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; }