首页 > 代码库 > poj1442 Black Box 栈和优先队列
poj1442 Black Box 栈和优先队列
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
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 43 1 -4 2 8 -1000 21 2 6 6Sample Output
3312
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 0x0f0f0f0f18 #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 }
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。