POJ 3481 SBT Basics

1 indicates that the customer K is inserted, the priority is P (equivalent to the size), and 2 indicates that the customer with the highest priority is output (that is, the maximum value is found) and deleted. 3. Similarly, the output level is the lowest. Idea: The get_min () and get_max () operations in SBT can be easily completed. For details, see [cpp] # include <iostream> # include <cstdio> # include <algorithm> # include <string> # include <cmath> # include <cstring> # include <queue> # include <set> # include <vector> # include <stack> # include <map> # include <iomanip> # define PI acos (-1.0) # define Max 100005 # define inf 1 <28 # define LL (x) (x <1) # define RR (x) (x <1 | 1) # define FOR (I, s, t) for (int I = (s); I <= (t); ++ I) # define ll long # define Mem (a, B) memset (a, B, sizeof (a) # define mp (a, B) make_pair (a, B) using namespace std; struct SBT {int left, right, num, size, priority;} tree [Max]; void left_rot (int & x) // left-hand {int y = tree [x]. right; tree [x]. right = tree [y]. left; tree [y]. left = x; tree [y]. size = tree [x]. size; tree [x]. size = tree [tree [x]. left]. size + tree [tree [x]. right]. size + 1; x = y;} void right_rot (int & x) // right-hand {int y = tree [x ]. Left; tree [x]. left = tree [y]. right; tree [y]. right = x; tree [y]. size = tree [x]. size; tree [x]. size = tree [tree [x]. left]. size + tree [tree [x]. right]. size + 1; x = y;} void maintain (int & x, bool flag) // update {if (! Flag) // The left subtree is 0 {if (tree [tree [x]. left]. left]. size> tree [tree [x]. right]. size) right_rot (x); else if (tree [tree [x]. left]. right]. size> tree [tree [x]. right]. size) {left_rot (tree [x]. left); right_rot (x);} else return;} else // The right subtree is 1 {if (tree [tree [tree [x]. right]. right]. size> tree [tree [x]. left]. size) left_rot (x); else if (tree [tree [x]. right]. left]. size> tree [tree [x]. left]. size) {right_rot (Tree [x]. right); left_rot (x);} else return;} maintain (tree [x]. left, 0); maintain (tree [x]. right, 1); maintain (x, 1); maintain (x, 0);} int root, top; void insert (int & x, int num, int priority) // insert here, p is the priority, that is, the size stored in the SBT, And that num can be understood as the customer's name and output only. {If (x = 0) {x = ++ top; tree [x]. size = 1; tree [x]. left = tree [x]. right = 0; tree [x]. num = num; tree [x]. priority = priority;} else {tree [x]. size ++; if (priority <tree [x]. priority) insert (tree [x]. left, num, priority); else insert (tree [x]. right, num, priority); maintain (root, priority> = tree [x]. priority) ;}} int del (int & x, int priority) {int d_priority; if (! X) return 0; tree [x]. size --; if (priority = tree [x]. priority | (tree [x]. priority> priority & tree [x]. left = 0) | (tree [x]. priority <priority & tree [x]. right = 0) {d_priority = tree [x]. priority; if (tree [x]. left & tree [x]. right) {tree [x]. priority = del (tree [x]. left, tree [x]. priority + 1);} else x = tree [x]. left + tree [x]. right;} else if (priority> tree [x]. priority) d_priority = del (tree [X]. right, priority); else if (priority <tree [x]. priority) d_priority = del (tree [x]. left, priority); return d_priority;} int get_min () {int x; for (x = root; tree [x]. left; x = tree [x]. left); printf ("% d \ n", tree [x]. num); return tree [x]. priority; // return this value, used to delete} int get_max () {int x; for (x = root; tree [x]. right; x = tree [x]. right); printf ("% d \ n", tree [x]. num); return tree [x]. priority; // similarly} Int main () {int n; root = top = 0; while (scanf ("% d", & n) {if (n = 1) {int a, B; scanf ("% d", & a, & B); insert (root, a, B);} else if (n = 2) {if (! Root) puts ("0"); else {int k = get_max (); del (root, k) ;}} else {if (! Root) puts ("0"); else {int k = get_min (); del (root, k) ;}} return 0 ;}