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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。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 | #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:\n", ++ 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\n", sbt.select(sz / 2 + 1)); } } } return 0;} |
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原文地址：http://www.cnblogs.com/jklongint/p/4548175.html
