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http://acm.hdu.edu.cn/showproblem.php?pid=5349
A simple problem
Problem Description
You have a multiple set,and now there are 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 empty now,the answer is 0)
The first line contains a number $N\ (N\leq 10^6)$,representing the number of operations.
Next $N$ line ,each line contains one or two numbers,describe one operation.
The number in this set is not greater than $10^9$.
For each operation 3,output a line representing the answer.
8
1 2
1 3
1 4
2
2
3
2
3
4
0
直接模拟即可,我用的平衡树。。
#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::pair; 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 *= 10) += 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->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->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; }
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原文地址:http://www.cnblogs.com/GadyPu/p/4796291.html