POJ1442 Black Box 优先队列(堆维护)+思维
2017-02-25 15:54
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1.题目描述:
Black Box
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.
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
Sample Output
Source
Northeastern Europe 1996
2.题意概述:
当时题目看了好一会,不容易懂,大概意思是给定M个数,每次可以插入序列一个数;再给N个数,表示在插入第几个数时输出一个数,第一次输出序列中最小的,第二次输出序列中第二小的……以此类推,直到输出N个数。
对于样例:
7 4
3 1 -4 2 8 -1000 2
1 2 6 6
7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。
3.解题思路:
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。我们可以用堆(优先队列)来做。
定义两个堆,一个用来存储前k小的数(大顶堆),大数在前,小数在后;另一个优先队列第k+1小到最大的数(小顶堆),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。
4.AC代码:
#include <iostream>
#include <stdio.h>
#include <queue>
#include <algorithm>
#include <functional>
#define maxn 303000
using namespace std;
typedef long long ll;
ll num[maxn];
int n, m;
priority_queue<ll>big;
priority_queue<ll, vector<ll>, greater<ll> >small;
int main()
{
int m, n;
scanf("%d%d", &m, &n);
for (int i = 1; i <= m; i++)
scanf("%lld", &num[i]);
int u, cnt = 1;
for (int i = 1; i <= n; i++)
{
scanf("%d", &u);
while (cnt <= u)
{
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
{
ll n1 = big.top();
ll n2 = small.top();
big.pop();
small.pop();
big.push(n2);
small.push(n1);
}
cnt++;
}
printf("%lld\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;
}
Black Box
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 12573 | Accepted: 5154 |
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
Source
Northeastern Europe 1996
2.题意概述:
当时题目看了好一会,不容易懂,大概意思是给定M个数,每次可以插入序列一个数;再给N个数,表示在插入第几个数时输出一个数,第一次输出序列中最小的,第二次输出序列中第二小的……以此类推,直到输出N个数。
对于样例:
7 4
3 1 -4 2 8 -1000 2
1 2 6 6
7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。
3.解题思路:
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。我们可以用堆(优先队列)来做。
定义两个堆,一个用来存储前k小的数(大顶堆),大数在前,小数在后;另一个优先队列第k+1小到最大的数(小顶堆),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。
4.AC代码:
#include <iostream>
#include <stdio.h>
#include <queue>
#include <algorithm>
#include <functional>
#define maxn 303000
using namespace std;
typedef long long ll;
ll num[maxn];
int n, m;
priority_queue<ll>big;
priority_queue<ll, vector<ll>, greater<ll> >small;
int main()
{
int m, n;
scanf("%d%d", &m, &n);
for (int i = 1; i <= m; i++)
scanf("%lld", &num[i]);
int u, cnt = 1;
for (int i = 1; i <= n; i++)
{
scanf("%d", &u);
while (cnt <= u)
{
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
{
ll n1 = big.top();
ll n2 = small.top();
big.pop();
small.pop();
big.push(n2);
small.push(n1);
}
cnt++;
}
printf("%lld\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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