题意：3种操作分别为入队，出队，查询当前队列的中位数。操作数为1e5数量级。
思路：先考虑离线算法，可以离散+线段树，可以划分树，考虑在线算法，则有treap名次树，SBtree(size balanced tree)等等。
///这个模板有问题，别再用了。。。！！！！
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#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <queue>using namespace std;const int maxn = 1e5 + 7;template<typename T> class SBTree {public: SBTree() { root = nil = new Node(); nil->ch[0] = nil->ch[1] = nil; nil->sz = 0; tot = top = 0; } ~SBTree() {} void clear() { root = nil; tot = top = 0; } int size() { return root->sz; } bool empty() { return root == nil; } void insert(const T &it) { insert(root, it); } void erase(const T &it) { erase(root, it); } bool find(const T &it) { return find(root, it); } const T &minItem() { return minMax(root, 0); } const T &maxItem() { return minMax(root, 1); } const T &select(int k) { return select(root, k); } int rank(const T &it) { return rank(root, it); }private: const static int maxn = 1e4 + 7; struct Node { T key; int sz; Node *ch[2]; } v[maxn], *stk[maxn], *root, *nil; int tot, top; void rotate(Node *&x, int d) { Node *y = x->ch[d ^ 1]; x->ch[d ^ 1] = y->ch[d]; y->ch[d] = x; y->sz = x->sz; x->sz = x->ch[0]->sz + x->ch[1]->sz + 1; x = y; } void maintain(Node *&x, int d) { if (x == nil) return; if (x->ch[d]->sz < x->ch[d ^ 1]->ch[d ^ 1]->sz) rotate(x, d); else if (x->ch[d]->sz < x->ch[d ^ 1]->ch[d]->sz) { rotate(x->ch[d ^ 1], d ^ 1); rotate(x, d); } else { return; } maintain(x->ch[0], 1); maintain(x->ch[1], 0); maintain(x, 1); maintain(x, 0); } void insert(Node *&x, const T &it) { if (x == nil) { x = top ? stk[top--] : v + tot++; x->key = it; x->sz = 1; x->ch[0] = x->ch[1] = nil; } else { ++x->sz; insert(x->ch[x->key < it], it); maintain(x, it < x->key); } } void erase(Node *&x, const T &it) { Node *p; if (x == nil) return; --x->sz; if (it < x->key) erase(x->ch[0], it); else if (x->key < it) erase(x->ch[1], it); else { if (x->ch[1] == nil) { stk[++top] = x; x = x->ch[0]; } else { for (p = x->ch[1]; p->ch[0] != nil; p = p->ch[0]); erase(x->ch[1], x->key = p->key); } } } bool find(const Node *x, const T &it) { if (x == nil || !(it < x->key || x->key < it)) return x != nil; return find(x->ch[x->key < it], it); } const T &minMax(const Node *x, int d) { return x->ch[d] == nil ? x->key : minMax(x->ch[d], d); } const T &select(const Node *x, int k) { if (x == nil || k == x->ch[0]->sz + 1) return x->key; return select(x->ch[x->ch[0]->sz + 1 < k], x->ch[0]->sz + 1 < k ? k - x->ch[0]->sz - 1 : k); } int rank(const Node *x, const T &it) { if (x == nil) return 1; if (it < x->key) return rank(x->ch[0], it); if (x->key < it) return rank(x->ch[1], it) + x->ch[0]->sz + 1; return x->ch[0]->sz + 1; }};SBTree<int> sbt;int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin);#endif // ONLINE_JUDGE int cas = 0, n; while (cin >> n) { printf("Case #%d:
", ++ cas); sbt.clear(); queue<int> Q; while (n --) { char s[10]; scanf("%s", s); if (s[0] == 'i') { int x; scanf("%d", &x); sbt.insert(x); Q.push(x); } else if (s[0] == 'o') { int x = Q.front(); Q.pop(); sbt.erase(x); } else { int sz = Q.size(); printf("%d
", sbt.select(sz / 2 + 1)); } } } return 0;} |