Hdu 5349 MZL ' s simple problem

Topic Connection

http://acm.hdu.edu.cn/showproblem.php?pid=5349

MZL ' s simple problemdescription

A Simple problem
Problem Description
You has a multiple set,and now there is three kinds of operations:
1 x:add number x to set
2:delete the minimum number (if the set is empty now,then ignore it)
3:query the maximum number (if the set is an empty now,the answer is 0)

Input

The first line contains a number $N \ (N\leq 10^6) $,representing the number of operations.
Next $N $ line, each line contains one or both Numbers,describe one operation.
The number in this set was not greater than $10^9$.

Output

For each operation 3,output a line representing the answer.

Sample Input

8
1 2
1 3
1 4
2
2
3
2
3

Sample Output

4
0

Direct simulation can be, I use the balance tree.

#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include < cstdio> #include <vector> #include <map>using std::map;using std::min;using std::find;using std::p air; Using std::vector;using std::multimap; #define ALL (c) (c). Begin (), (c). End () #define ITER (c) Decltype ((c). Begin ()) #  Define Cpresent (C, E) (Find (All (c), (e))! = (c). End ()) #define REP (i, n) for (int i = 0; i < (int) (n); i++) #define TR (c, i) for (ITER (c) i = (c). Begin (); I! = (c). end (); ++i) #define PB (E) push_back (e) #define MP (A, b) Make_pair (A, b) const int    N = 1 << 20;const int INF = 0x3f3f3f3f;inline int read () {char c; int R;    while (((c = GetChar ()) < ' 0 ' | | c > ' 9 ') && C ^ '-'); bool F = c = = '-'; if (f) r = 0;    else R = C-' 0 ';    while ((c = GetChar ()) >= ' 0 ' && C <= ' 9 ') (r *=) + = C-' 0 '; if (f) return-r; else return R;}    struct Node {int V, S, c;    Node *ch[2];     inline void Set (int _v, int _s, Node *p) {   v = _v, s = c = _s;    Ch[0] = ch[1] = p;    } inline void Push_up () {s = ch[0]->s + Ch[1]->s + C;    } inline int cmp (int x) const {return x = = v? -1:x > v;    }};struct sizebalancetree {int top;    Node *null, *root, *tail;    Node Stack[n], *pool[n >> 1];        inline void init () {top = 0;        Tail = &stack[0];        NULL = tail++;        Null->set (0, 0, NULL);    root = null;        } inline Node *newnode (int v) {Node *x =!top? tail++: Pool[--top];        X->set (V, 1, null);    return x;         } inline void Rotate (node *&x, int D) {node *k = x->ch[!d]; X->ch[!d] = K->ch[d], k->ch[d] = x; K->s = x->s; X->push_up ();    x = k;        } inline void maintain (Node *&x, int d) {if (!x->s) return;        if (X->ch[d]->ch[d]->s > X->ch[!d]->s) rotate (x,!d); else if (X->ch[d]->ch[!d]->s > X->ch[!d]->s) rotate (X-&GT;CH[D], D), rotate (x,!d);        else return;    Maintain (x, 0), maintain (x, 1);        } inline void Insert (Node *&x, int v) {if (!x->s) {x = NewNode (v); return;}        x->s++;        int d = x->cmp (v);        if ( -1 = = d) {x->c++; return;}        Insert (X->ch[d], V);        X->push_up ();    Maintain (x, d);        } inline void Erase (Node *&x, int p) {if (!x->s) return;        x->s--;        int d = X-&GT;CMP (p);            if ( -1 = = d) {if (X->c > 1) {x->c--; return;}                else if (!x->ch[0]->s | |!x->ch[1]->s) {pool[top++] = x; x = X->ch[0]->s?            X->ch[0]: x->ch[1];                } else {Node *ret = x->ch[1];                for (; ret->ch[0]->s; ret = ret->ch[0]);            Erase (x->ch[1], x->v = ret->v);        }} else {erase (x->ch[d], p); } if (X->s) X->pusH_up ();    } inline void Insert (int v) {insert (root, v);    } inline void Erase () {Erase (Root, kth (1));        } inline int kth (int k) {Node *x = root; for (int t = 0; x->s;)            {T = x->ch[0]->s;            if (k <= t) x = x->ch[0];            else if (t + 1 <= k && k <= T + x->c) break;        else K-= t + x->c, x = x->ch[1];    } return x->v;        } inline void query () {if (!root->s) {puts ("0"); return;}    printf ("%d\n", KTH (Root->s));    }}sbt;int Main () {#ifdef LOCAL freopen ("In.txt", "R", stdin);    Freopen ("OUT.txt", "w+", stdout); #endif sbt.init ();    int N, v, op;        while (~SCANF ("%d", &n)) {sbt.init ();            Rep (i, n) {op = read ();            if (1 = = op) v = Read (), Sbt.insert (v);            if (2 = = op) sbt.erase ();        if (3 = = op) sbt.query (); }} return 0;}

Hdu 5349 MZL ' s simple problem