【POJ1442】【Treap】Black Box
| Description
Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:
ADD (x): put element x into Black Box; GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending. Let us examine a possible sequence of 11 transactions: Example 1
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type. Let us describe the sequence of transactions by two integer arrays: 1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2). 2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6). The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence. Input
Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.
Output
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.
Sample Input 7 4 3 1 -4 2 8 -1000 2 1 2 6 6 Sample Output 3 3 1 2 Source |
【分析】
练下手而已，没什么。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <vector> 6 #include <utility> 7 #include <iomanip> 8 #include <string> 9 #include <cmath> 10 #include <queue> 11 #include <assert.h> 12 #include <map> 13 14 const int N = 30000 + 10; 15 const int SIZE = 250;//块状链表的大小 16 const int M = 50000 + 5; 17 using namespace std; 18 struct TREAP{ 19 struct Node{ 20 int fix, size; 21 int val; 22 Node *ch[2]; 23 }mem[30000 + 10], *root; 24 int tot; 25 //大随机 26 int BIG_RAND(){return (rand() * RAND_MAX + rand());} 27 Node *NEW(){ 28 Node *p = &mem[tot++]; 29 p->fix = BIG_RAND(); 30 p->val = 0; 31 p->size = 1; 32 p->ch[0] = p->ch[1] = NULL; 33 return p; 34 } 35 //将t的d节点换到t 36 void rotate(Node *&t, int d){ 37 Node *p = t->ch[d]; 38 t->ch[d] = p->ch[d ^ 1]; 39 p->ch[d ^ 1] = t; 40 t->size = 1; 41 if (t->ch[0] != NULL) t->size += t->ch[0]->size; 42 if (t->ch[1] != NULL) t->size += t->ch[1]->size; 43 t = p; 44 t->size = 1; 45 if (t->ch[0] != NULL) t->size += t->ch[0]->size; 46 if (t->ch[1] != NULL) t->size += t->ch[1]->size; 47 return; 48 } 49 void insert(Node *&t, int val){ 50 //插入 51 if (t == NULL){ 52 t = NEW(); 53 t->val = val; 54 return; 55 } 56 //大的在右边,小的在左边 57 int dir = (val >= t->val); 58 insert(t->ch[dir], val); 59 //维护最大堆的性质 60 if (t->ch[dir]->fix > t->fix) rotate(t, dir); 61 t->size = 1; 62 if (t->ch[0] != NULL) t->size += t->ch[0]->size; 63 if (t->ch[1] != NULL) t->size += t->ch[1]->size; 64 } 65 //在t的子树中找到第k小的值 66 int find(Node *t, int k){ 67 if (t->size == 1) return t->val; 68 int l = 0;//t的左子树中有多少值 69 if (t->ch[0] != NULL) l += t->ch[0]->size; 70 if (k == (l + 1)) return t->val; 71 if (k <= l) return find(t->ch[0], k); 72 else return find(t->ch[1], k - (l + 1)); 73 } 74 }treap; 75 typedef long long ll; 76 int have[N];//have为1则在这个地方GET 77 int data[N], m, n; 78 79 void init(){ 80 treap.root = NULL; 81 treap.tot = 0; 82 memset(have, 0, sizeof(have)); 83 scanf("%d%d", &m, &n); 84 for (int i = 1; i <= m; i++) scanf("%d", &data[i]); 85 for (int i = 1; i <= n; i++){ 86 int x; 87 scanf("%d", &x); 88 have[x]++; 89 } 90 } 91 void work(){ 92 int pos = 0;//代表要获得的位置 93 for (int i = 1; i <= m; i++){ 94 treap.insert(treap.root, data[i]); 95 //printf("%d", treap.root->size); 96 while (have[i]){ 97 pos++; 98 printf("%d\n", treap.find(treap.root, pos)); 99 have[i]--; 100 } 101 } 102 103 } 104 105 int main(){ 106 int T; 107 #ifdef LOCAL 108 freopen("data.txt", "r", stdin); 109 freopen("out.txt", "w", stdout); 110 #endif 111 init(); 112 work(); 113 return 0; 114 }