BZOJ2286 [Sdoi2011消耗战

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\n", 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 }
View Code

(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)

posted on 2015-01-27 21:19  Xs酱~  阅读(602)  评论(0)  编辑  收藏  举报