poj1442 treap

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poj 1442

题意就是给定了一些数要插入，插入的时候顺便问当前第i小的元素是多少。

http://poj.org/problem?id=1442

用treap写了这个题。

下面的代码以后就当做模版使用了。

基本上和大白书的一样。一些修改的地方是网上看其他的网友的抄过来的。

#include <cstdio>#include <cstring>#include <cstdlib>#include <ctime>using namespace std;struct Node{Node *ch[2]; //左右子树int r;  //优先级，数值越大，优先级越高int v; // 键值int s;  //size域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;}}elseremove(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));//随机种子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)); //查询当前第x大元素}//removetree(rt);}return 0;}

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