题意：有n个数，按顺序加入，求加入前Gi个数时第i个最小的数是多少
思路：这里需要用到STL里的优先队列priority_queue，建一个大堆和一个小堆，若想在一个无序的序列里找第n个小的数，可以先把一个序列的n-1个数放入大堆（即假设这n-1个数是该序列里最小的），然后向小堆里push数，若小堆头元素<大堆头元素（即最小的元素比最大的元素大，说明大堆里的数还不是最小的），则交换两个数。难点：这里如果只用一个小堆的话会超时；另外每输入的一个m数，要查询的最小的数正好是第i++个，即这里每输出一个数就把该数存入大堆里，正好使大堆里数的个数不断++（由0个不断增加），以方便下一次查询第i++个小的。 Black Box
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 7205 | Accepted: 2930 |
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 6
Sample Output
3 3 1 2
1 #include <stdio.h>
2 #include <string.h>
3 #include <iostream>
4 #include <algorithm>
5 #include <cstdio>
6 #include <cstring>
7 #include <cmath>
8 #include <stack>
9 #include <queue>
10 #include <functional>
11 #include <vector>
12 #include <map>
13 //priority_queue<int> pq;
14 //queue<int >q;
15 using namespace std;
16
17 #define M 0x0f0f0f0f
18 #define min(a,b) (a>b?b:a)
19 #define max(a,b) (a>b?a:b)
20 int main()
21 {
22 int m,n;
23 int i,j,b,c,t;
24 int a[30005];
25 scanf("%d %d",&n,&m);
26 priority_queue< int,vector<int>,greater<int> >small;
27 priority_queue< int>large;
28 for(i=0; i<n; i++)
29 scanf("%d",&a[i]);
30 c=0;
31 for(i=0; i<m; i++)
32 {
33 scanf("%d",&b);
34 while(c<b)
35 {
36 small.push(a[c]);
37 if(!large.empty()&&small.top()<large.top())
38 {
39 t=small.top();
40 small.pop();
41 small.push(large.top());
42 large.pop();
43 large.push(t);
44 }
45 c++;
46 }
47 printf("%d\n",small.top());
48 //精华！！！！
49 large.push(small.top());//每次i++,让large里放一个数，使得以后每次比较时
50 small.pop(); //保证small里的第一个数是第i个最小的！！！！！
51 }
52 return 0;
53 }
poj1442 Black Box 栈和优先队列