poj3481——treap

来源：互联网发布：sql数据库实训心得编辑：程序博客网时间：2024/05/21 03:57
题意：动态求一个集合的最大值和最小值。
treap的模板题。
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;struct node{    node * ch[2];    int r, v, q;    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;    }};node * root;void rotate(node * & o, int d) // d = 0 左旋 d = 1 右旋{    node * k = o->ch[d ^ 1]; o->ch[d ^ 1] = k->ch[d]; k->ch[d] = o; o = k;}void insert(node * & o, int x, int q){    if(o == NULL) { o = new node(); o->ch[0] = o->ch[1] = NULL; o->v = x; o->r = rand(); o->q = q; }    else    {        int d = o->cmp(x);        insert(o->ch[d], x, q);        if(o->ch[d] > o) rotate(o, d ^ 1);    }}void remove(node * & o, int x){    int d = o->cmp(x);    if(d == -1)    {        if(o->ch[0] == NULL) { o = o->ch[1]; }        else if(o ->ch[1] == NULL) { o = o->ch[0]; }        else        {            int d2 = (o->ch[0] > o->ch[1]) ? 1 : 0;            rotate(o, d2); remove(o->ch[d2], x);        }    }    else remove(o->ch[d], x);}int findmax(node * & o){    if(o->ch[1] == NULL) { int q = o->q; remove(o, o->v); return q; }    else return findmax(o->ch[1]);}int findmin(node * & o){    if(o->ch[0] == NULL) { int q = o->q; remove(o, o->v); return q; }    else return findmin(o->ch[0]);}int main(){    freopen("in", "r", stdin);    root = NULL;    int cas;    while(~scanf("%d", &cas) && cas != 0)    {        if(cas == 1)        {            int k, p; scanf("%d %d", &k, &p);            insert(root, p, k);        }        else if(cas == 2)        {            if(root == NULL) { printf("0\n"); continue; }            printf("%d\n", findmax(root));        }        else        {            if(root == NULL) { printf("0\n"); continue; }            printf("%d\n", findmin(root));        }    }    return 0;}/*21 20 141 30 321 10 99322002030100*/