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 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 6Sample Output
3 3 1 2题解: 用最大堆保存每次取出的最小数,然后用当前最小堆队列的最小值和最大堆的比较,如果大于最大堆需要两个值互换, 应该最大堆的top保存的是第k-1小的数,如果最小堆有比它小的,那么最大堆的top应变成第k小的数。 #include<iostream> #include<algorithm> #include<cmath> #include<queue> #include<cstring> using namespace std; const int maxn = 30100; int num[maxn]; priority_queue<int> big; priority_queue<int,vector<int>,greater<int> > small; int main(){ int n,m,op; scanf("%d%d",&n,&m); for(int i=0;i<n;i++){ scanf("%d",&num[i]); } int cnt = 0; for(int i=0;i<m;i++){ scanf("%d",&op); while(cnt<op){ small.push(num[cnt]); if(!big.empty()&&big.top()>small.top()){ int tmp1 = big.top(); int tmp2 = small.top(); big.pop(); small.pop(); big.push(tmp2); small.push(tmp1); } cnt++; } printf("%d\n",small.top()); big.push(small.top()); small.pop(); } return 0; }
转载于:https://www.cnblogs.com/yuanshixingdan/p/6080940.html
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