[ACM] POJ 1442 Black Box (堆,优先队列)
2018-02-28 18:25
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Black Box
DescriptionOur 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
4000
<= 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.
InputInput 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.OutputWrite to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input7 4
3 1 -4 2 8 -1000 2
1 2 6 6Sample Output3
3
1
2SourceNortheastern Europe 1996
解题思路:题意不好懂。。我直接解释下样例什么意思把。7 4
3 1 -4 2 8 -1000 2
1 2 6 67代表下面给定7个数的数字序列,4可以理解为四次查询, 1 2 6 6为查讯,第一次查询是求数字序列只有前一个数(3)时,此时的第一小的数字,即1,第二次查询是求数字序列只有前2个数(3 1)时,此时的第二小的数字,即 3,第三次查询是求数字序列只有前6个数时(3,1,-4,2,8,-1000),此时的第三小的数字,即1,第四次查询是求数字序列只有前6个数时,此时的第四小的数字。思路为: 当求第4小的数字时,我们用一个大顶堆来维护前三个最小的数字,那么小顶堆的堆顶即为所求。用priority_queue<int>big; 优先队列还起到大顶堆的作用,顶部即为最大值 ,即 big.top();
priority_queue<int,vector<int>,greater<int> >small; 小顶堆,顶堆为最小值 ,即 small.top();用样例来说明一下 大顶堆和小顶堆工作方法:1 2 6 6首先是1: 把第一个数3 插入到小顶堆中,这时候判断,如果大顶堆不为空且小顶堆的top小于大顶堆的top时,就把二者的top值互换,因为,大顶堆中的数不能比小顶堆中的大,大顶堆维护第i次查询时,前i-1个最小的数,这时候大顶堆为空,不用互换值, 输出第一次查询时前1个数的第1小的数即为 小顶堆的top 3,然后把小顶堆的top 移除,放到大顶堆中。然后是2: 把第二个数1插入到小顶堆中,判断,大顶堆不为空,小顶.top() 1 <大顶.top() 3 ,所以二者互换,小顶.top()为3,大顶top()为1, 然后输出小顶.top(),即为第二小的数。把小顶的.top()移除,放入大顶中。这是大顶中维护的是目前数字中前两个最小的数。然后是6: 要求前6个数中第3小的数字,先得把3 1以后的四个数字插入才能够六个数,依次插入, -4插入小顶堆,这时 大顶堆 3 1,-4<3,互换, 小顶堆 3,大顶堆为1 -4 ,2插入小顶堆,小顶堆 2,3 大顶堆 1 -4, 2>1,不用互换, 8插入到小顶堆,小顶堆为 2,3,8 大顶堆为 1,-4, 2>1,不用互换,-1000插入到小顶堆,小顶堆 -1000,2,3,8,大顶堆1,-4, -1000<1,不行,互换以后得小顶堆,1,2,3,8 , 大顶堆 -1000,-4 ,输出小顶堆.top() 1, 并把它移除放入到大顶堆中,为下次查询做准备。所以从以上可以看出,关键点就是第i次查询时,大顶堆中维护的总是目前数字中最小的 i-1个数。
代码:[cpp] view plain copy#include <iostream>
#include <stdio.h>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn=30010;
int num[maxn];
int n,m;
int main()
{
priority_queue<int>big;
priority_queue<int,vector<int>,greater<int> >small;
int m,n;
scanf("%d%d",&m,&n);
for(int i=1;i<=m;i++)
scanf("%d",&num[i]);
int cnt=1;
int op;
for(int i=1;i<=n;i++)
{
cin>>op;
while(cnt<=op)
{
small.push(num[cnt]);
if(!big.empty()&&small.top()<big.top())//小顶堆里面的数不能比大顶堆里面的数小
{
int n1=big.top();
int n2=small.top();
big.pop();
small.pop();
big.push(n2);
small.push(n1);
}
cnt++;
}
printf("%d\n",small.top());
big.push(small.top());//这句话很关键,保证了在求第k个最小数时,大顶堆里面保存的是前k-1个最小数
small.pop();
}
return 0;
}
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 7099 | Accepted: 2888 |
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
4000
<= 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.
InputInput 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.OutputWrite to the output Black Box answers sequence for a given sequence of transactions, one number each line.Sample Input7 4
3 1 -4 2 8 -1000 2
1 2 6 6Sample Output3
3
1
2SourceNortheastern Europe 1996
解题思路:题意不好懂。。我直接解释下样例什么意思把。7 4
3 1 -4 2 8 -1000 2
1 2 6 67代表下面给定7个数的数字序列,4可以理解为四次查询, 1 2 6 6为查讯,第一次查询是求数字序列只有前一个数(3)时,此时的第一小的数字,即1,第二次查询是求数字序列只有前2个数(3 1)时,此时的第二小的数字,即 3,第三次查询是求数字序列只有前6个数时(3,1,-4,2,8,-1000),此时的第三小的数字,即1,第四次查询是求数字序列只有前6个数时,此时的第四小的数字。思路为: 当求第4小的数字时,我们用一个大顶堆来维护前三个最小的数字,那么小顶堆的堆顶即为所求。用priority_queue<int>big; 优先队列还起到大顶堆的作用,顶部即为最大值 ,即 big.top();
priority_queue<int,vector<int>,greater<int> >small; 小顶堆,顶堆为最小值 ,即 small.top();用样例来说明一下 大顶堆和小顶堆工作方法:1 2 6 6首先是1: 把第一个数3 插入到小顶堆中,这时候判断,如果大顶堆不为空且小顶堆的top小于大顶堆的top时,就把二者的top值互换,因为,大顶堆中的数不能比小顶堆中的大,大顶堆维护第i次查询时,前i-1个最小的数,这时候大顶堆为空,不用互换值, 输出第一次查询时前1个数的第1小的数即为 小顶堆的top 3,然后把小顶堆的top 移除,放到大顶堆中。然后是2: 把第二个数1插入到小顶堆中,判断,大顶堆不为空,小顶.top() 1 <大顶.top() 3 ,所以二者互换,小顶.top()为3,大顶top()为1, 然后输出小顶.top(),即为第二小的数。把小顶的.top()移除,放入大顶中。这是大顶中维护的是目前数字中前两个最小的数。然后是6: 要求前6个数中第3小的数字,先得把3 1以后的四个数字插入才能够六个数,依次插入, -4插入小顶堆,这时 大顶堆 3 1,-4<3,互换, 小顶堆 3,大顶堆为1 -4 ,2插入小顶堆,小顶堆 2,3 大顶堆 1 -4, 2>1,不用互换, 8插入到小顶堆,小顶堆为 2,3,8 大顶堆为 1,-4, 2>1,不用互换,-1000插入到小顶堆,小顶堆 -1000,2,3,8,大顶堆1,-4, -1000<1,不行,互换以后得小顶堆,1,2,3,8 , 大顶堆 -1000,-4 ,输出小顶堆.top() 1, 并把它移除放入到大顶堆中,为下次查询做准备。所以从以上可以看出,关键点就是第i次查询时,大顶堆中维护的总是目前数字中最小的 i-1个数。
代码:[cpp] view plain copy#include <iostream>
#include <stdio.h>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn=30010;
int num[maxn];
int n,m;
int main()
{
priority_queue<int>big;
priority_queue<int,vector<int>,greater<int> >small;
int m,n;
scanf("%d%d",&m,&n);
for(int i=1;i<=m;i++)
scanf("%d",&num[i]);
int cnt=1;
int op;
for(int i=1;i<=n;i++)
{
cin>>op;
while(cnt<=op)
{
small.push(num[cnt]);
if(!big.empty()&&small.top()<big.top())//小顶堆里面的数不能比大顶堆里面的数小
{
int n1=big.top();
int n2=small.top();
big.pop();
small.pop();
big.push(n2);
small.push(n1);
}
cnt++;
}
printf("%d\n",small.top());
big.push(small.top());//这句话很关键,保证了在求第k个最小数时,大顶堆里面保存的是前k-1个最小数
small.pop();
}
return 0;
}
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