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http://acm.hdu.edu.cn/showproblem.php?pid=4288
In mathematics and computer science, an algorithm describes a set of procedures or instructions that define a procedure. The term has become increasing popular since the advent of cheap and reliable computers. Many companies now employ a single coder to write an algorithm that will replace many other employees. An added benefit to the employer is that the coder will also become redundant once their work is done. 1
You are now the signle coder, and have been assigned a new task writing code, since your boss would like to replace many other employees (and you when you become redundant once your task is complete).
Your code should be able to complete a task to replace these employees who do nothing all day but eating: make the digest sum.
By saying “digest sum” we study some properties of data. For the sake of simplicity, our data is a set of integers. Your code should give response to following operations:
1. add x – add the element x to the set;
2. del x – remove the element x from the set;
3. sum – find the digest sum of the set. The digest sum should be understood by
\begin{align*}\large{\sum_{1 \leq i \leq k}{a_i}}\ \ \ \ \ \ \ {where \ \ i\ \ mod \ \ 5\ = \ 3}\end{align*}
where the set S is written as $\large{a_1, a_2, ... , a_k}$ satisfying $\large{a_1 < a_2 < a_3 < ... < a_k}$
There’re several test cases.
In each test case, the first line contains one integer $N ( 1 \leq N \leq 10^5 )$, the number of operations to process.
Then following is n lines, each one containing one of three operations: “add x” or “del x” or “sum”.
You may assume that $1 \leq x \leq 10^9.$
Please see the sample for detailed format.
For any “add x” it is guaranteed that x is not currently in the set just before this operation.
For any “del x” it is guaranteed that x must currently be in the set just before this operation.
Please process until EOF (End Of File).
For each operation “sum” please print one line containing exactly one integer denoting the digest sum of the current set. Print 0 if the set is empty.
9
add 1
add 2
add 3
add 4
add 5
sum
add 6
del 3
sum
6
add 1
add 3
add 5
add 7
add 9
sum
3
4
5
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<vector> 7 #include<map> 8 using std::cin; 9 using std::cout; 10 using std::endl; 11 using std::find; 12 using std::map; 13 using std::pair; 14 using std::vector; 15 #define all(c) (c).begin(), (c).end() 16 #define iter(c) decltype((c).begin()) 17 #define cpresent(c, e) (find(all(c), (e)) != (c).end()) 18 #define rep(i, n) for (int i = 0; i < (int)(n); i++) 19 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i) 20 #define pb(e) push_back(e) 21 #define mp(a, b) make_pair(a, b) 22 const int Max_N = 110000; 23 typedef int key_t; 24 typedef unsigned long long ull; 25 struct Node { 26 key_t key; 27 Node *ch[2]; 28 double fix; 29 int size; 30 ull sum[5]; 31 bool vis; 32 void set(const key_t &k, double pri, Node *p) { 33 key = k, fix = pri, size = 1; 34 vis = false; 35 ch[0] = ch[1] = p; 36 rep(i, 5) sum[i] = 0; 37 } 38 inline int cmp(const key_t v) const { 39 return v == key ? -1 : v > key; 40 } 41 }; 42 struct Treap { 43 int top; 44 Node *tail, *root, *null; 45 Node stack[Max_N], *pool[Max_N]; 46 inline double get_fix() { 47 return rand() / (double)RAND_MAX; 48 } 49 inline void init() { 50 top = 0; 51 tail = &stack[0]; 52 null = tail++; 53 null->set(0, 1.0, NULL); 54 null->size = 0; 55 root = null; 56 } 57 inline Node *newNode(const key_t &k, double pri) { 58 Node *p = null; 59 if (!top) p = tail++; 60 else p = pool[--top]; 61 p->set(k, pri, null); 62 return p; 63 } 64 inline void rotate(Node *&x, int d) { 65 Node *k = x->ch[!d]; 66 x->ch[!d] = k->ch[d]; 67 k->ch[d] = x; 68 x->vis = k->vis = false; 69 x = k; 70 } 71 inline void insert(Node *&x, const key_t &k, double p) { 72 if (x == null) { x = newNode(k, p); return; } 73 int d = x->cmp(k); 74 if (-1 == d) { x->vis = false; return; } 75 insert(x->ch[d], k, p); 76 if (x->fix > x->ch[d]->fix) rotate(x, !d); 77 x->vis = false; 78 return; 79 } 80 inline void erase(Node *&x, const key_t &k) { 81 if (!x->size) return; 82 int d = x->cmp(k); 83 if (-1 == d) { 84 if (!x->ch[0]->size || !x->ch[1]->size) { 85 pool[top++] = x; 86 x = x->ch[0]->size ? x->ch[0] : x->ch[1]; 87 } else { 88 int c = x->ch[0]->fix < x->ch[1]->fix; 89 rotate(x, !c); 90 erase(x->ch[!c], k); 91 } 92 } else { 93 erase(x->ch[d], k); 94 } 95 if (x->size) x->vis = false; 96 return; 97 } 98 inline void insert(const key_t &k) { 99 insert(root, k, get_fix()); 100 } 101 inline void erase(const key_t &k) { 102 erase(root, k); 103 } 104 inline int count(Node *x) { 105 return !x->size ? 0 : x->size; 106 } 107 inline void dfs(Node *x) { 108 if (!x->size) return; 109 if (!x->vis) { 110 Node *lch = x->ch[0], *rch = x->ch[1]; 111 dfs(lch); dfs(rch); 112 int lcnt = count(lch); 113 int rcnt = count(rch); 114 x->size = lcnt + 1 + rcnt; 115 rep(i, 5) x->sum[i] = 0; 116 if (lch->size) rep(i, 5) x->sum[i] = lch->sum[i]; 117 x->sum[lcnt % 5] += x->key; 118 if (rch->size) rep(i, 5) x->sum[(lcnt + 1 + i) % 5] += rch->sum[i]; 119 x->vis = true; 120 } 121 } 122 inline void go() { 123 int x; 124 char buf[256]; 125 scanf("%s", buf); 126 if (buf[0] == ‘a‘) { 127 scanf("%d", &x); 128 insert(x); 129 } else if (buf[0] == ‘d‘) { 130 scanf("%d", &x); 131 erase(root, x); 132 } else { 133 dfs(root); 134 if (!root->size) puts("0"); 135 else printf("%lld\n", root->sum[2]); 136 } 137 } 138 }trp; 139 int main() { 140 #ifdef LOCAL 141 freopen("in.txt", "r", stdin); 142 freopen("out.txt", "w+", stdout); 143 #endif 144 int Q; 145 while (~scanf("%d", &Q)) { 146 trp.init(); 147 rep(q, Q) trp.go(); 148 } 149 return 0; 150 }
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原文地址：http://www.cnblogs.com/GadyPu/p/4566545.html
