• POJ2104 K-th Number


    某谷链接


    发一篇博客记录一下主席树板子。
    别忘了输出原数组中的元素,而不是离散化后的。

    #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 = 1e5 + 5;
    const int maxt = 2e6 + 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], b[maxn];
    
    struct Tree
    {
      int ls, rs, sum;
    }t[maxt];
    int root[maxn], cnt = 0;
    
    void insert(int old, int& now, int l, int r, int d)
    {
      t[now = ++cnt] = t[old];
      t[now].sum++;
      if(l == r) return;
      int mid = (l + r) >> 1;
      if(d <= mid) insert(t[old].ls, t[now].ls, l, mid, d);
      else insert(t[old].rs, t[now].rs, mid + 1, r, d);
    }
    int query(int old, int now, int l, int r, int id)
    {
      if(l == r) return l;
      int mid = (l + r) >> 1, del = t[t[now].ls].sum - t[t[old].ls].sum;
      if(id <= del) return query(t[old].ls, t[now].ls, l, mid, id);
      else return query(t[old].rs, t[now].rs, mid + 1, r, id - del);
    }
    
    int main()
    {
      n = read(); m = read();
      for(int i = 1; i <= n; ++i) a[i] = b[i] = read();
      sort(b + 1, b + n + 1);
      int _n = unique(b + 1, b + n + 1) - b - 1;
      for(int i = 1; i <= n; ++i)
        a[i] = lower_bound(b + 1, b + _n + 1, a[i]) - b;
      for(int i = 1; i <= n; ++i) insert(root[i - 1], root[i], 1, _n, a[i]);
      for(int i = 1, L, R, k; i <= m; ++i)
        {
          L = read(); R = read(); k = read();
          write(b[query(root[L - 1], root[R], 1, _n, k)]), enter;
        }
      return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/mrclr/p/10065164.html
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