一眼，这不是裸树形dp嘛~
一阵猛敲，敲完发现多组询问。。。额，不会了。。。
围观hzwer，发现这就是虚树嘛！咦、等等，虚树是什么？就是个神奇的东西啦！
构建虚树复杂度是O(k * log(k))，dp的复杂度都是O(k)级别的，由于Σki <= 50W，所以就行了。
构建方法是单调性dp，感觉好神啊！
1 /************************************************************** 2 Problem: 2286 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:6148 ms 7 Memory:38896 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 typedef long long ll; 15 const int N = 250005; 16 const int M = N << 1; 17 const ll inf = (ll) 1e60; 18 19 struct edge { 20 int next, to, v; 21 edge() {} 22 edge(int _n, int _t, int _v) : next(_n), to(_t), v(_v) {} 23 } e[M]; 24 25 struct Edge { 26 int next, to; 27 Edge() {} 28 Edge(int _n, int _t) : next(_n), to(_t) {} 29 } E[M]; 30 31 int first[N], First[N], tot; 32 33 struct tree_node { 34 int fa[19], dep, w; 35 ll mn; 36 } tr[N]; 37 38 int n, m, cnt_dfs; 39 int s[N], top; 40 ll f[N]; 41 int K; 42 int h[N], cnt_h; 43 44 int read() { 45 int x = 0; 46 char ch = getchar(); 47 while (ch < '0' || '9' < ch) 48 ch = getchar(); 49 while ('0' <= ch && ch <= '9') 50 (x *= 10) += ch - '0', ch = getchar(); 51 return x; 52 } 53 54 inline bool cmp(int a, int b) { 55 return tr[a].w < tr[b].w; 56 } 57 58 inline void Add_Edges(int x, int y, int z) { 59 e[++tot] = edge(first[x], y, z), first[x] = tot; 60 e[++tot] = edge(first[y], x, z), first[y] = tot; 61 } 62 63 inline void add_edge(int x, int y) { 64 if (x == y) return; 65 E[++tot] = Edge(First[x], y), First[x] = tot; 66 } 67 68 #define y e[x].to 69 void dfs(int p) { 70 int x; 71 tr[p].w = ++cnt_dfs; 72 for (x = 1; x <= 18; ++x) 73 tr[p].fa[x] = tr[tr[p].fa[x - 1]].fa[x - 1]; 74 for (x = first[p]; x; x = e[x].next) 75 if (y != tr[p].fa[0]) { 76 tr[y].mn = min(tr[p].mn, (ll) e[x].v); 77 tr[y].dep = tr[p].dep + 1; 78 tr[y].fa[0] = p; 79 dfs(y); 80 } 81 } 82 #undef y 83 84 inline int get_lca(int x, int y) { 85 int i; 86 if (tr[x].dep < tr[y].dep) swap(x, y); 87 for (i = 18; ~i; --i) 88 if (tr[tr[x].fa[i]].dep >= tr[y].dep) x = tr[x].fa[i]; 89 if (x == y) return x; 90 for (i = 18; ~i; --i) 91 if (tr[x].fa[i] != tr[y].fa[i]) 92 x = tr[x].fa[i], y = tr[y].fa[i]; 93 return tr[x].fa[0]; 94 } 95 96 #define y E[x].to 97 void dp(int p) { 98 ll tmp = 0; 99 int x; 100 for (x = First[p]; x; x = E[x].next) 101 dp(y), tmp += f[y]; 102 First[p] = 0; 103 f[p] = (tmp != 0 && tmp <= tr[p].mn ? tmp : tr[p].mn); 104 } 105 #undef y 106 107 void work() { 108 int i, now, lca; 109 K = read(), tot = 0; 110 for (i = 1; i <= K; ++i) h[i] = read(); 111 sort(h + 1, h + K + 1, cmp); 112 for (cnt_h = 1, i = 2; i <= K; ++i) 113 if (get_lca(h[cnt_h], h[i]) != h[cnt_h]) h[++cnt_h] = h[i]; 114 s[top = 1] = 1; 115 for (i = 1; i <= cnt_h; ++i) { 116 now = h[i], lca = get_lca(now, s[top]); 117 while (1) { 118 if (tr[lca].dep >= tr[s[top - 1]].dep) { 119 add_edge(lca, s[top--]); 120 if (s[top] != lca) s[++top] = lca; 121 break; 122 } 123 add_edge(s[top - 1], s[top]); 124 --top; 125 } 126 if (s[top] != now) s[++top] = now; 127 } 128 while (--top) add_edge(s[top], s[top + 1]); 129 dp(1); 130 printf("%lld ", f[1]); 131 } 132 133 int main() { 134 int i, x, y, z; 135 n = read(); 136 for (i = 1; i < n; ++i) { 137 x = read(), y = read(), z = read(); 138 Add_Edges(x, y, z); 139 } 140 tr[1].mn = inf, tr[1].dep = 1; 141 dfs(1); 142 m = read(); 143 while (m--) work(); 144 return 0; 145 }
(p.s. 700次submit留念O(∩_∩)O)
(p.s.1 感谢卢爷提出复杂度分析的错误Orzzzzzz)