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http://acm.hdu.edu.cn/showproblem.php?pid=5249
你工作以后, KPI 就是你的全部了. 我开发了一个服务，取得了很大的知名度。数十亿的请求被推到一个大管道后同时服务从管头拉取请求。让我们来定义每个请求都有一个重要值。我的KPI是由当前管道内请求的重要值的中间值来计算。现在给你服务记录，有时我想知道当前管道内请求的重要值得中间值。
有大约100组数据。
每组数据第一行有一个n(1≤n≤10000)，代表服务记录数。
接下来有n行，每一行有3种形式
"in x": 代表重要值为x(0≤x≤109)的请求被推进管道。
"out": 代表服务拉取了管道头部的请求。
"query: 代表我想知道当前管道内请求重要值的中间值. 那就是说，如果当前管道内有m条请求, 我想知道，升序排序后第$floor(m/2)+1_{th}$ 条请求的重要值.
为了让题目简单，所有的x都不同，并且如果管道内没有值，就不会有"out"和"query"操作。
对于每组数据，先输出一行
Case #i:
然后每一次"query"，输出当前管道内重要值的中间值。
6
in 874
query
out
in 24622
in 12194
query
Case #1:
874
24622
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<queue> 7 using std::queue; 8 const int Max_N = 10010; 9 struct Node { 10 int data, s; 11 bool color; 12 Node *fa, *ch[2]; 13 inline void set(int _v, int i, bool _color, Node *p) { 14 data = _v, color = _color, s = i; 15 fa = ch[0] = ch[1] = p; 16 } 17 inline void push_up() { 18 s = ch[0]->s + ch[1]->s + 1; 19 } 20 inline void push_down() { 21 for (Node *x = this; x->s; x = x->fa) x->s--; 22 } 23 }; 24 struct RedBlackTree { 25 int top; 26 Node *root, *null; 27 Node stack[Max_N], *tail, *store[Max_N]; 28 void init() { 29 tail = &stack[0]; 30 null = tail++; 31 null->set(0, 0, 0, NULL); 32 root = null; 33 top = 0; 34 } 35 inline Node *newNode(int v) { 36 Node *p = null; 37 if (!top) p = tail++; 38 else p = store[--top]; 39 p->set(v, 1, 1, null); 40 return p; 41 } 42 inline void rotate(Node* &x, bool d) { 43 Node *y = x->ch[!d]; 44 x->ch[!d] = y->ch[d]; 45 if (y->ch[d] != null) y->ch[d]->fa = x; 46 y->fa = x->fa; 47 if (x->fa == null) root = y; 48 else x->fa->ch[x->fa->ch[0] != x] = y; 49 y->ch[d] = x; 50 x->fa = y; 51 y->s = x->s; 52 x->push_up(); 53 } 54 inline void insert(int v) { 55 Node *x = root, *y = null; 56 while (x->s){ 57 x->s++; 58 y = x, x = x->ch[v > x->data]; 59 } 60 x = newNode(v); 61 if (y != null) y->ch[v > y->data] = x; 62 else root = x; 63 x->fa = y; 64 insert_fix(x); 65 } 66 inline void insert_fix(Node* &x) { 67 while (x->fa->color){ 68 Node *par = x->fa, *Gp = par->fa; 69 bool d = par == Gp->ch[0]; 70 Node *uncle = Gp->ch[d]; 71 if (uncle->color) { 72 par->color = uncle->color = 0; 73 Gp->color = 1; 74 x = Gp; 75 } else if (x == par->ch[d]) { 76 rotate(x = par, !d); 77 } else { 78 Gp->color = 1; 79 par->color = 0; 80 rotate(Gp, d); 81 } 82 } 83 root->color = 0; 84 } 85 inline Node *find(Node *x, int data) { 86 while (x->s && x->data != data) x = x->ch[x->data < data]; 87 return x; 88 } 89 inline void del_fix(Node* &x) { 90 while (x != root && !x->color) { 91 bool d = x == x->fa->ch[0]; 92 Node *par = x->fa, *sibling = par->ch[d]; 93 if (sibling->color) { 94 sibling->color = 0; 95 par->color = 1; 96 rotate(x->fa, !d); 97 sibling = par->ch[d]; 98 } else if (!sibling->ch[0]->color && !sibling->ch[1]->color) { 99 sibling->color = 1, x = par; 100 } else { 101 if (!sibling->ch[d]->color) { 102 sibling->ch[!d]->color = 0; 103 sibling->color = 1; 104 rotate(sibling, d); 105 sibling = par->ch[d]; 106 } 107 sibling->color = par->color; 108 sibling->ch[d]->color = par->color = 0; 109 rotate(par, !d); 110 break; 111 } 112 } 113 x->color = 0; 114 } 115 inline void del(int data) { 116 Node *z = find(root, data); 117 if (!z->s) return; 118 Node *y = z, *x = null; 119 if (z->ch[0]->s && z->ch[1]->s) { 120 y = z->ch[1]; 121 while (y->ch[0]->s) y = y->ch[0]; 122 } 123 x = y->ch[!y->ch[0]->s]; 124 x->fa = y->fa; 125 if (!y->fa->s) root = x; 126 else y->fa->ch[y->fa->ch[1] == y] = x; 127 if (z != y) z->data = y->data; 128 y->fa->push_down(); 129 if (!y->color) del_fix(x); 130 store[top++] = y; 131 } 132 inline int kth(int k) { 133 int t = 0; 134 Node *x = root; 135 for (; x->s;){ 136 t = x->ch[0]->s; 137 if (k == t + 1) break; 138 else if (k <= t) x = x->ch[0]; 139 else k -= t + 1, x = x->ch[1]; 140 } 141 return x->data; 142 } 143 int operator[] (int k) { 144 return kth(k); 145 } 146 }rbt; 147 int main(){ 148 #ifdef LOCAL 149 freopen("in.txt", "r", stdin); 150 freopen("out.txt", "w+", stdout); 151 #endif 152 int n, v, c = 1; 153 char buf[100]; 154 while (~scanf("%d", &n)) { 155 rbt.init(); queue<int> q; 156 printf("Case #%d:\n", c++); 157 while (n--) { 158 scanf("%s", buf); 159 if (‘i‘ == buf[0]) { 160 scanf("%d", &v); 161 rbt.insert(v), q.push(v); 162 } else if (‘o‘ == buf[0]) { 163 v = q.front(); q.pop(); 164 rbt.del(v); 165 } else { 166 printf("%d\n", rbt[((int)q.size() >> 1) + 1]); 167 } 168 } 169 } 170 return 0; 171 }
1 #include<iostream> 2 #include<cstdlib> 3 #include<cstring> 4 #include<cstdio> 5 #include<queue> 6 using std::queue; 7 const int Max_N = 10010; 8 struct Node { 9 int v, s; 10 Node *ch[2]; 11 inline void set(int _v, int _s, Node *p) { 12 v = _v, s = _s; 13 ch[0] = ch[1] = p; 14 } 15 inline void push_up() { 16 s = ch[0]->s + ch[1]->s + 1; 17 } 18 inline int cmp(int x) const { 19 return x == v ? -1 : x > v; 20 } 21 }; 22 struct SizeBalanceTree { 23 Node stack[Max_N]; 24 Node *root, *null, *tail; 25 Node *store[Max_N]; 26 int top; 27 void init() { 28 tail = &stack[0]; 29 null = tail++; 30 null->set(0, 0, NULL); 31 root = null; 32 top = 0; 33 } 34 inline Node *newNode(int v) { 35 Node *p = null; 36 if (top) p = store[--top]; 37 else p = tail++; 38 p->set(v, 1, null); 39 return p; 40 } 41 inline void rotate(Node* &x, int d) { 42 Node *k = x->ch[!d]; 43 x->ch[!d] = k->ch[d]; 44 k->ch[d] = x; 45 k->s = x->s; 46 x->push_up(); 47 x = k; 48 } 49 inline void Maintain(Node* &x, int d) { 50 if (x->ch[d] == null) return; 51 if (x->ch[d]->ch[d]->s > x->ch[!d]->s) { 52 rotate(x, !d); 53 } else if (x->ch[d]->ch[!d]->s > x->ch[!d]->s) { 54 rotate(x->ch[d], d), rotate(x, !d); 55 } else { 56 return; 57 } 58 Maintain(x, 0), Maintain(x, 1); 59 } 60 inline void insert(Node* &x, int v) { 61 if (x == null) { 62 x = newNode(v); 63 return; 64 } else { 65 x->s++; 66 int d = x->cmp(v); 67 insert(x->ch[d], v); 68 x->push_up(); 69 Maintain(x, d); 70 } 71 } 72 inline void del(Node* &x, int v) { 73 if (!x->s) return; 74 x->s--; 75 int d = x->cmp(v); 76 if (-1 == d) { 77 if (!x->ch[0]->s || !x->ch[1]->s) { 78 store[top++] = x; 79 x = x->ch[0]->s ? x->ch[0] : x->ch[1]; 80 } else { 81 Node *ret = x->ch[1]; 82 for (; ret->ch[0] != null; ret = ret->ch[0]); 83 del(x->ch[1], x->v = ret->v); 84 } 85 } else { 86 del(x->ch[d], v); 87 } 88 if (x->s) x->push_up(); 89 } 90 inline void insert(int v) { 91 insert(root, v); 92 } 93 inline void del(int v) { 94 del(root, v); 95 } 96 inline int kth(int k) { 97 int t; 98 Node *x = root; 99 for (; x->s;) { 100 t = x->ch[0]->s; 101 if (k <= t) x = x->ch[0]; 102 else if (t + 1 == k) break; 103 else k -= t + 1, x = x->ch[1]; 104 } 105 return x->v; 106 } 107 int operator[] (int k) { 108 return kth(k); 109 } 110 }sbt; 111 int main(){ 112 #ifdef LOCAL 113 freopen("in.txt", "r", stdin); 114 freopen("out.txt", "w+", stdout); 115 #endif 116 int n, v, c = 1; 117 char buf[100]; 118 while (~scanf("%d", &n)) { 119 sbt.init(); queue<int> q; 120 printf("Case #%d:\n", c++); 121 while (n--) { 122 scanf("%s", buf); 123 if (‘i‘ == buf[0]) { 124 scanf("%d", &v); 125 sbt.insert(v), q.push(v); 126 } else if (‘o‘ == buf[0]) { 127 v = q.front(); q.pop(); 128 sbt.del(v); 129 } else { 130 printf("%d\n", sbt[((int)q.size() >> 1) + 1]); 131 } 132 } 133 } 134 return 0; 135 }
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原文地址：http://www.cnblogs.com/GadyPu/p/4542487.html
