POJ 1442 Black Box [Treap]

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Description

给两个数组
A
U
A表示插入到集合的数，U表示在插入第U[i]的数后询问
询问第K大的数
K每次询问之后加一

Algorithm

Treap实现的名次树的模板题
看算法竞赛入门经典训练指南

Code

#include <iostream>#include <cstdlib>#include <cstdio>using namespace std;struct Node {        Node *ch[2]; //左右子树 0左1右        int r; //优先级        int v; //值        int s; //子树        Node(int v) : v(v) {                ch[0] = ch[1] = NULL;                r = rand();                s = 1;        }        int cmp(int x) {                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, bool d){        Node *k = o->ch[!d];        o->ch[!d] = k->ch[d];        k->ch[d] = o;        o->maintain();        k->maintain();        o = k;}void insert(Node *&o, int x) //在 o 节点插入值为 x的节点{        if (o == NULL) {                o = new Node(x);        } else {                bool d = x > (o->v);                insert(o->ch[d], x);                if (o->ch[d]->r > o->r) rotate(o, !d);        }        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) {                        bool d2 = o->ch[0]->r < o->ch[1]->r;                        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);                }        }}int n, m;const int MAXM = 30000 + 9;const int MAXN = MAXM;int a[MAXM];int u[MAXN];void solve(){        for (int i = 0; i < m; i++) {                scanf("%d", &a[i]);        }        for (int i = 0; i < n; i++) {                scanf("%d", &u[i]);        }        int j = 0;        Node *treap = NULL;        for (int i = 0; i < n; i++) {                for (; j < u[i]; j++) {                        insert(treap, a[j]);                }                printf("%d\n", kth(treap, i + 1));        }}int main(){        //freopen("in.txt", "r", stdin);        while (scanf("%d%d", &m, &n) != EOF) {                solve();        }}

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