支配(灭绝)树学习笔记 暨 P2597 [ZJOI2012]灾难 题解

支配树:满足树上任何一点的所有祖先都是他的支配点的树.

支配点:在图中指定一个起点s,若删去一个点t,s到w就没有路径,则t为w的支配点.

• 一棵树本身就是支配树

• 对于DAG:

  • 在原图上拓扑排序.
  • 按照拓扑序建支配树,方法是求出所有原图上有边指向该节点的点(即反图上该节点指向的点)的支配树上lca,将该节点设为lca的子结点,顺便更新深度,fa数组等信息.
  • dfs求出每个点的子树size.
  • 结果即为每个点子树(不含该点)的size.

• 对于一般图:

  • 去问tarjan.

• #include <iostream>
  #include <cstdio>
  #include <queue>
  using namespace std;
  const int N = 100005;
  int n, x;
  int rin[N], head[3][N * 2], cnt[3], index[N], dep[N], fa[N][21], siz[N];
  queue<int> q;
  struct edge
  {
      int to, next;
  } e[3][N * 2];
  void add(int u, int v, int type)
  {
      e[type][++cnt[type]].to = v;
      e[type][cnt[type]].next = head[type][u];
      head[type][u] = cnt[type];
  }
  void topo()
  {
      for (int i = 1; i <= n; i++)
          if (!rin[i])
              q.push(i);
      int tot = 0;
      while (!q.empty())
      {
          int x = q.front();
          q.pop();
          index[++tot] = x;
          for (int i = head[0][x]; i; i = e[0][i].next)
          {
              int v = e[0][i].to;
              rin[v]--;
              if (!rin[v])
                  q.push(v);
          }
      }
  }
  int lca(int a, int b)
  {
      if (dep[a] > dep[b])
          swap(a, b);
      for (int i = 20; i >= 0; i--)
          if (dep[fa[b][i]] >= dep[a]) 
              b = fa[b][i];
      if (a == b)
          return a;
      for (int i = 20; i >= 0; i--)
          if (fa[a][i] != fa[b][i])
              a = fa[a][i], b = fa[b][i];
      return fa[a][0];
  }
  void dfs(int k)
  {
      for (int i = head[2][k]; i; i = e[2][i].next)
      {
          int v = e[2][i].to;
          dfs(v);
          siz[k] += siz[v];
      }
      siz[k]++;
  }
  int main()
  {
      ios::sync_with_stdio(false);
      cin >> n;
      for (int i = 1; i <= n; i++)
      {
          while (1)
          {
              cin >> x;
              if (x == 0)
                  break;
              rin[i]++;
              add(x, i, 0);
              add(i, x, 1);
          }
      }
      topo();
      for (int i = 1; i <= n; i++)
      {
          int ans = e[1][head[1][index[i]]].to;
          for (int j = head[1][index[i]]; j; j = e[1][j].next)
              ans = lca(ans, e[1][j].to);
          add(ans, index[i], 2);
          dep[index[i]] = dep[ans] + 1;
          fa[index[i]][0] = ans;
          for (int j = 1; j <= 20; j++)
              fa[index[i]][j] = fa[fa[index[i]][j - 1]][j - 1];
      }
      dfs(0);
      for (int i = 1; i <= n; i++)
          cout << siz[i] - 1<<endl;
  }
  

标签：支配,head,int,题解,cnt,fa,P2597,ZJOI2012,type
来源： https://www.cnblogs.com/KinuhataSaiai/p/15555374.html