[SDOI2011]消耗战

嘟嘟嘟


今天学了个虚树，觉得这东西挺奇特。


虚树简单来说就是保留每一次询问的关键点以及他们的lca，把其他的点从树中删去。这样如果每一次询问的复杂度为(O(n))的话，总复杂度就是(O(sum _ {i = 1} ^ {m} k_i))。
虚树的构建方法这里推fjzzq大佬的博客。


构建完虚树后，在虚树上dp。因为每一个点按dfs序排好了，所以可以省去递归，直接倒着dp，用每一个点的dp值更新他的父亲。


还有就是每一次只把用过的点清空，memset会超时。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 2.5e5 + 5;;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m, a[maxn];
struct Edge
{
  int nxt, to, w;
}e[maxn << 1];
int head[maxn], ecnt = -1;
void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], y, w};
  head[x] = ecnt;
}

int dep[maxn], dfsx[maxn], cnt = 0;
int fa[21][maxn], Min[21][maxn];
inline void dfs(const int& now, const int& _f)
{
  dfsx[now] = ++cnt;
  for(int i = 1; (1 << i) <= dep[now]; ++i)
    {
      fa[i][now] = fa[i - 1][fa[i - 1][now]];
      Min[i][now] = min(Min[i - 1][now], Min[i - 1][fa[i - 1][now]]);
    }
  for(int i = head[now], v; i != -1; i = e[i].nxt)
    {
      if((v = e[i].to) == _f) continue;
      dep[v] = dep[now] + 1;
      fa[0][v] = now; Min[0][v] = e[i].w;
      dfs(v, now);
    }
}
inline int LCA(int x, int y)
{
  if(dep[x] < dep[y]) swap(x, y);
  for(int i = 20; i >= 0; --i)
    if(dep[x] - (1 << i) >= dep[y]) x = fa[i][x];
  if(x == y) return x;
  for(int i = 20; i >= 0; --i)
    if(fa[i][x] != fa[i][y]) x = fa[i][x], y = fa[i][y];
  return fa[0][x];
}

int vir[maxn];  //虚树点集
int st[maxn], top = 0;
int pa[maxn];
bool cmp(int x, int y) {return dfsx[x] < dfsx[y];}
inline void build(int& m)
{
  vir[++m] = 1; top = 0;
  sort(vir + 1, vir + m + 1, cmp);
  int tp = m;
  for(int i = 1; i <= tp; ++i)
    {
      int now = vir[i];
      if(!top) {pa[st[++top] = now] = 0; continue;}
      int lca = LCA(st[top], now);
      while(dep[st[top]] > dep[lca])
	{
	  if(dep[st[top - 1]] < dep[lca]) pa[st[top]] = lca;
	  --top;
	}
      if(lca != st[top])
	{
	  vir[++m] = lca;
	  pa[lca] = st[top];
	  st[++top] = lca;
	}
      pa[st[++top] = now] = lca; 
    }
  sort(vir + 2, vir + m + 1, cmp);
}

inline int dis(int x, int y)  //两点间最小边权
{
  int ret = INF;
  if(dep[x] < dep[y]) swap(x, y);
  for(int i = 20; i >= 0; --i)
    if(dep[x] - (1 << i) >= dep[y]) ret = min(ret, Min[i][x]), x = fa[i][x];
  if(x == y) return ret;
  for(int i = 20; i >= 0; --i)
    if(fa[i][x] != fa[i][y])
      {
	ret = min(ret, min(Min[i][x], Min[i][y]));
	x = fa[i][x], y = fa[i][y];
      }
  return min(ret, min(Min[0][x], Min[0][y]));
}
bool vis[maxn];
int wi[maxn];
ll dp[maxn];
inline ll solve(const int& m)
{
  for(int i = 2; i <= m; ++i) wi[vir[i]] = dis(vir[i], pa[vir[i]]);
  for(int i = 1; i <= m; ++i) dp[vir[i]] = 0;
  for(int i = m, v; i > 1; --i)
    {
      if(vis[v = vir[i]]) dp[pa[v]] += wi[v];
      else dp[pa[v]] += min(dp[v], (ll)wi[v]);
    }
  return dp[1];
}  

int main()
{
  Mem(head, -1);
  n = read();
  for(int i = 1; i < n; ++i)
    {
      int x = read(), y = read(), w = read();
      addEdge(x, y, w); addEdge(y, x, w);
    }
  dfs(1, 0);
  int T = read();
  while(T--)
    {
      m = read();
      for(int i = 1; i <= m; ++i) vis[vir[i] = read()] = 1;
      build(m);
      write(solve(m)), enter;
      for(int i = 1; i <= m; ++i) vis[vir[i]] = 0;
    }
  return 0;
}