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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; Let us examine a possible sequence of 11 transactions: Example 1 N Transaction i Black Box contents after transaction Answer
(elements are arranged by non-descending)
1 ADD(3) 0 3
2 GET 1 3 3
3 ADD(1) 1 1, 3
4 GET 2 1, 3 3
5 ADD(-4) 2 -4, 1, 3
6 ADD(2) 2 -4, 1, 2, 3
7 ADD(8) 2 -4, 1, 2, 3, 8
8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8
9 GET 3 -1000, -4, 1, 2, 3, 8 1
10 GET 4 -1000, -4, 1, 2, 3, 8 2
11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
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 }