Black Box
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 11150 | Accepted: 4554 |
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.
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
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
心累到现在都没过……
分析：因为输出时是按照先输出最小的，再输出第二小这样的方式输出的，相当于依次输出一个有序序列中的值。但因为这个序列不是固定不变的，而是不断的在更新，所以用数组是无法实现的。我们可以用优先队列来做。
定义两个优先队列，一个用来存储前k小的数，大数在前，小数在后；另一个优先队列第k+1小到最大的数，小数在前，大数在后。每次拿到一个数，先判断第一个优先队列中的数满不满k个，如果不满k个，则直接把这个数压入到第一个队列；如果满k个，判断这个数和第一个优先队列中的第一个数的大小：如果比第一个数大，就压入第二个优先队列；如果比第一个数小，就把第一个优先队列的队首元素弹出压入第二个队列，把这个新数压入第一个优先队列。
输出时，如果第一个优先队列里的元素个数小于k，则先把第二个优先队列里的队首元素弹出压入第一个优先队列，然后输出第一个优先队列的队首元素；如果满k个，则直接输出第一个优先队列的队首元素。
1 #include "bits/stdc++.h" 2 #define mem(a,b) memset(a,b,sizeof(a)) 3 using namespace std; 4 typedef long long LL; 5 const int MAX=30005; 6 int n,m; 7 int a[MAX],r[MAX]; 8 struct Que{ 9 int h[MAX*10]; 10 int n; 11 void ini(){mem(h,0),n=0;} 12 void heapify1(int x){ 13 int child=x*2,key=h[x]; 14 while (child<=n){ 15 if (child<n && h[child]<h[child+1]) child++; 16 if (key<h[child]){h[x]=h[child];x=child;child=x*2;} 17 else break; 18 } 19 h[x]=key; 20 } 21 void insert1(int key){ 22 int x=++n; 23 while (x>1){ 24 if (key>h[x/2]) h[x]=h[x/2],x/=2; 25 else break; 26 } 27 h[x]=key; 28 } 29 void del1(){ 30 if (n==1) n=0; 31 else h[1]=h[n--],heapify1(1); 32 } 33 void heapify2(int x){ 34 int child=x*2,key=h[x]; 35 while (child<=n){ 36 if (child<n && h[child]>h[child+1]) child++; 37 if (h[x]>h[child]){h[x]=h[child];x=child;child=x*2;} 38 else break; 39 } 40 h[x]=key; 41 } 42 void insert2(int key){ 43 int x=++n; 44 while (x>1){ 45 if (key<h[x/2]) h[x]=h[x/2],x/=2; 46 else break; 47 } 48 h[x]=key; 49 } 50 void del2(){ 51 if (n==1) n=0; 52 else h[1]=h[n--],heapify2(1); 53 } 54 }q1,q2; 55 int main(){ 56 freopen ("box.in","r",stdin); 57 freopen ("box.out","w",stdout); 58 int i,j,k; 59 while (~scanf("%d%d",&n,&m)){ 60 q1.ini();q2.ini(); 61 for (i=1;i<=n;i++) 62 scanf("%d",a+i); 63 for (j=1;j<=m;j++) 64 scanf("%d",r+j); 65 sort(r+1,r+m+1); 66 int index(1); 67 for (i=1;i<=n;i++){ 68 if (index>n) 69 break; 70 if (q1.n<index) 71 q1.insert1(a[i]); 72 else if (a[i]<q1.h[1]){ 73 int zt=q1.h[1]; 74 q1.del1(); 75 q1.insert1(a[i]); 76 q2.insert2(zt); 77 } 78 else 79 q2.insert2(a[i]); 80 while (i==r[index]){ 81 printf("%d ",q1.h[1]); 82 if (q2.n>0){ 83 q1.insert1(q2.h[1]); 84 q2.del2(); 85 } 86 index++; 87 } 88 } 89 } 90 return 0; 91 }