[BZOJ4530] [BJOI 2014] 대융합.

[제목 링크]
  • 클릭하여 링크 열기
  • [발상 요점]
  • LinkCutTree로 이 나무를 유지하고 가벼운 아버지에게 자수 정보를 기록하면 된다.
  • 시간 복잡도 O(QLogN) O(Q L o g N).

  • 【코드】
    
    #include
    
    using namespace std;
    const int MAXN = 100005;
    template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
    template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
    template <typename T> void read(T &x) {
      x = 0; int f = 1;
      char c = getchar();
      for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
      for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
      x *= f;
    }
    template <typename T> void write(T x) {
      if (x < 0) x = -x, putchar('-');
      if (x > 9) write(x / 10);
      putchar(x % 10 + '0');
    }
    template <typename T> void writeln(T x) {
      write(x);
      puts("");
    }
    struct LinkCutTree {
      struct Node {
          int child[2];
          int father, up;
          int size, light;
          bool rev;
      } a[MAXN];
      int size, n;
      void init(int x) {
          n = x;
          for (int i = 1; i <= n; i++)
              a[i].size = 1;
      }
      void update(int root) {
          a[root].size = a[root].light + 1;
          if (a[root].child[0]) a[root].size += a[a[root].child[0]].size;
          if (a[root].child[1]) a[root].size += a[a[root].child[1]].size;
      }
      void pushdown(int root) {
          if (a[root].rev) {
              swap(a[root].child[0], a[root].child[1]);
              if (a[root].child[0]) a[a[root].child[0]].rev ^= true;
              if (a[root].child[1]) a[a[root].child[1]].rev ^= true;
              a[root].rev = false;
          }
      }
      bool get(int x) {
          return a[a[x].father].child[1] == x;
      }
      void rotate(int x) {
          int f = a[x].father, g = a[f].father;
          a[x].up = a[f].up, a[f].up = 0;
          pushdown(f), pushdown(x);
          bool tmp = get(x);
          a[f].child[tmp] = a[x].child[tmp ^ 1];
          if (a[f].child[tmp]) a[a[f].child[tmp]].father = f;
          a[f].father = x;
          a[x].child[tmp ^ 1] = f;
          a[x].father = g;
          if (g) a[g].child[a[g].child[1] == f] = x;
          update(f), update(x);
      }
      void splay(int x) {
          pushdown(x);
          for (int f = a[x].father; (f = a[x].father) != 0; rotate(x))
              if (a[f].father) {
                  if (get(x) == get(f)) rotate(f);
                  else rotate(x);
              }
      }
      void access(int x) {
          splay(x);
          int tmp = a[x].child[1];
          if (tmp != 0) {
              a[tmp].up = x;
              a[tmp].father = 0;
              a[x].child[1] = 0;
              a[x].light += a[tmp].size;
              update(x);
          }
          while (a[x].up) {
              int f = a[x].up;
              splay(f);
              tmp = a[f].child[1];
              if (tmp != 0) {
                  a[tmp].up = f;
                  a[tmp].father = 0;
                  a[f].light += a[tmp].size;
              }
              a[f].child[1] = x;
              a[f].light -= a[x].size;
              a[x].father = f;
              a[x].up = 0;
              update(f);
              splay(x);
          }
      }
      void link(int x, int y) {
          access(x);
          splay(x);
          reverse(y);
          access(y);
          splay(y);
          a[x].child[1] = y;
          a[y].father = x;
          update(x);
      }
      void reverse(int x) {
          access(x);
          splay(x);
          a[x].rev ^= true;
      }
      long long query(int x, int y) {
          reverse(x);
          access(y);
          splay(x);
          int tmp = a[x].size;
          int tnp = a[y].size;
          return 1ll * (tmp - tnp) * tnp;
      }
    } LCT;
    int main() {
      int n, m;
      read(n), read(m);
      LCT.init(n);
      for (int i = 1; i <= m; i++) {
          char opt = getchar();
          while (opt != 'A' && opt != 'Q') opt = getchar();
          int x, y; read(x), read(y);
          if (opt == 'A') LCT.link(x, y);
          else writeln(LCT.query(x, y));
      }
      return 0;
    }

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