题目来源：POJ 1442 Black Box

题意：输入xi 输出前xi个数的第i大的数

思路：试了下自己的treap模版

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       using namespace std; struct Node { Node *ch[2]; int r; int v; int s; Node(){} Node(int v): v(v) { ch[0] = ch[1] = NULL; r = rand(); s = 1; } bool operator < (const Node& rhs) const{ return r < rhs.r; } int cmp(int x) const{ if(x == v) return -1; return x < v ? 0 : 1; } void maintain(){ s = 1; if(ch[0] != NULL) s += ch[0]->s; if(ch[1] != NULL) s += ch[1]->s; } }; void rotate(Node* &o, int d){ Node* k = o->ch[d^1]; o->ch[d^1] = k->ch[d]; k->ch[d] = o; o->maintain(); k->maintain(); o = k; } void insert(Node* &o, int x){ if(o == NULL){ o = new Node(x); } else{ int d = (x < o->v ? 0 : 1); insert(o->ch[d], x); if((o->ch[d]->r) > (o->r)) rotate(o, d^1); } o->maintain(); } void remove(Node* &o, int x){ int d= o->cmp(x); if(d == -1){ Node* u = o; if(o->ch[0] != NULL && o->ch[1] != NULL){ int d2 = (o->ch[0] > o->ch[1] ? 1 : 0); rotate(o, d2); remove(o->ch[d2], x); } else{ if(o->ch[0] == NULL) o = o->ch[1]; else o = o->ch[0]; delete u; } } else remove(o->ch[d], x); if(o != NULL) o->maintain(); } int kth(Node* o, int k){ if(o == NULL || k <= 0 || k > o->s) return 0; int s = (o->ch[0] == NULL ? 0 : o->ch[0]->s); if(k == s+1) return o->v; else if(k <= s) return kth(o->ch[0], k); else return kth(o->ch[1], k-s-1); } void removetree(Node* &x){ if(x->ch[0] != NULL) removetree(x->ch[0]); if(x->ch[1] != NULL) removetree(x->ch[1]); delete x; x = NULL; } int n, m, a[30010]; Node *rt = NULL; int main() { while(scanf("%d %d", &n, &m) != EOF) { srand(time(0)); for(int i = 1; i <= n; i++) scanf("%d", &a[i]); int l = 1; for(int i = 1; i <= m; i++) { int x; scanf("%d", &x); while(l <= x) { insert(rt, a[l]); l++; } printf("%d\n", kth(rt, i)); } //removetree(rt); } return 0; }