• Count on a tree II [树分块]


    bzoj2589
    大概是选几个关键点,然后跳来跳去就好了。
    b[40][40] 表示一条链上跳的QwQ

    // powered by c++11
    // by Isaunoya
    #include <bits/stdc++.h>
    #define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
    #define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
    using namespace std;
    using db = double;
    using ll = long long;
    using uint = unsigned int;
    #define Tp template
    using pii = pair<int, int>;
    #define fir first
    #define sec second
    Tp<class T> void cmax(T& x, const T& y) {
      if (x < y) x = y;
    }
    Tp<class T> void cmin(T& x, const T& y) {
      if (x > y) x = y;
    }
    #define all(v) v.begin(), v.end()
    #define sz(v) ((int)v.size())
    #define pb emplace_back
    Tp<class T> void sort(vector<T>& v) { sort(all(v)); }
    Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
    Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
    const int SZ = 1 << 23 | 233;
    struct FILEIN {
      char qwq[SZ], *S = qwq, *T = qwq, ch;
    #ifdef __WIN64
    #define GETC getchar
    #else
      char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
    #endif
      Tp<class T> void read(T& x) {
        bool sign = 0;
        while ((ch = GETC()) < 48) sign ^= (ch == 45);
        x = (ch ^ 48);
        while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48);
        x = sign ? -x : x;
      }
      FILEIN& operator>>(int& x) { return read(x), *this; }
    } in;
    struct FILEOUT {
      const static int LIMIT = 1 << 22;
      char quq[SZ], ST[233];
      int sz, O;
      ~FILEOUT() { flush(); }
      void flush() {
        fwrite(quq, 1, O, stdout);
        fflush(stdout);
        O = 0;
      }
      FILEOUT& operator<<(char c) { return quq[O++] = c, *this; }
      Tp<class T> void write(T x) {
        if (O > LIMIT) flush();
        if (x < 0) {
          quq[O++] = 45;
          x = -x;
        }
        do {
          ST[++sz] = x % 10 ^ 48;
          x /= 10;
        } while (x);
        while (sz) quq[O++] = ST[sz--];
      }
      FILEOUT& operator<<(int x) { return write(x), *this; }
    } out;
    
    int n, m;
    const int maxn = 4e4 + 2;
    bitset<maxn> qwq[42][42], qaq;
    vector<int> g[maxn];
    int a[maxn], b[maxn], sz[maxn], son[maxn], mxd[maxn], dep[maxn], fa[maxn];
    int id[maxn], cnt = 0;
    
    void dfs(int u) {
      sz[u] = 1, mxd[u] = dep[u];
      for (int v : g[u])
        if (!dep[v]) {
          dep[v] = dep[u] + 1, fa[v] = u;
          dfs(v), sz[u] += sz[v];
          cmax(mxd[u], mxd[v]);
          if (sz[v] > sz[son[u]]) son[u] = v;
        }
      if (mxd[u] - dep[u] >= 1000) id[u] = ++cnt, mxd[u] = dep[u];
    }
    
    int ff[maxn], st[maxn], top = 0;
    
    void dfs2(int u) {
      for (int v : g[u])
        if (dep[v] > dep[u]) {
          if (id[v]) {
            int ip = id[st[top]], in = id[v];
            for (int x = v; x != st[top]; x = fa[x]) qwq[ip][in].set(a[x]);
            qaq = qwq[ip][in];
            for (int i = 1; i < top; i++) {
            	qwq[id[st[i]]][in] = qwq[id[st[i]]][ip];
    					qwq[id[st[i]]][in] |= qaq;
    				}
            ff[v] = st[top], st[++top] = v;
          }
          dfs2(v);
          if (id[v]) --top;
        }
    }
    
    int tp[maxn];
    void dfs3(int u, int t) {
      tp[u] = t;
      if (son[u]) dfs3(son[u], t);
      for (int v : g[u])
        if (!tp[v]) dfs3(v, v);
    }
    
    int lca(int x, int y) {
      while (tp[x] != tp[y]) (dep[tp[x]] > dep[tp[y]]) ? x = fa[tp[x]] : y = fa[tp[y]];
      return dep[x] < dep[y] ? x : y;
    }
    
    void qwqwq(int x, int lca) {
      while (x != lca && !id[x]) qaq.set(a[x]), x = fa[x];
      if (x != lca) {
        int pre = x;
        while (dep[ff[pre]] >= dep[lca]) pre = ff[pre];
        if (pre != x) qaq |= qwq[id[pre]][id[x]];
        while (pre != lca) qaq.set(a[pre]), pre = fa[pre];
      }
    }
    
    int qry(int x, int y) {
      int Lca = lca(x, y);
      qaq.reset(), qaq.set(a[Lca]);
      qwqwq(x, Lca), qwqwq(y, Lca);
      return qaq.count();
    }
    
    signed main() {
      // code begin.
      in >> n >> m;
      rep(i, 1, n) in >> a[i], b[i] = a[i];
      sort(b + 1, b + n + 1);
      int len = unique(b + 1, b + n + 1) - b - 1;
      rep(i, 1, n) a[i] = lower_bound(b + 1, b + len + 1, a[i]) - b;
      rep(i, 2, n) {
        int u, v;
        in >> u >> v, g[u].pb(v), g[v].pb(u);
      }
      dfs(dep[1] = 1);
      if (!id[1]) id[1] = ++cnt;
      st[++top] = 1, dfs2(1), dfs3(1, 1);
      int ans = 0;
      while (m--) {
        int u, v;
        in >> u >> v, u ^= ans;
        out << (ans = qry(u, v)) << '
    ';
      }
      return 0;
      // code end.
    }
    
  • 相关阅读:
    Spring Cloud Alibaba(一)为什么选用Spring Cloud Alibaba?
    Spring Cloud Alibaba(三)Nacos配置中心
    配置JNDI数据源
    vue动态表单DynamicForm 吴小明
    css变量的声明和读取 吴小明
    【图片预览】第二种方式:hooks子组件调用父组件方法 吴小明
    react+antd快捷菜单搜索组件:MenuSearch 吴小明
    【图片预览】第一种方式:hooks父组件调用子组件方法(1、子组件中使用useImperativeHandle钩子 2、父组件中使用useRef)【reactviewer】 吴小明
    ios遮罩层滚动穿透问题 吴小明
    【图片预览】第三种方式:将组件数据放在公共状态下 吴小明
  • 原文地址:https://www.cnblogs.com/Isaunoya/p/12445910.html
Copyright © 2020-2023  润新知