poj 2823 单调队列

Sliding Window

Time Limit: 12000MS		Memory Limit: 65536K
Total Submissions: 559348		Accepted: 12040
Case Time Limit: 5000MS

Description

An array of size n ≤ 10⁶ is given to you. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves rightwards by one position. Following is an example:
The array is [1 3 -1 -3 5 3 6 7], and k is 3.

Window position	Minimum value	Maximum value
[1  3  -1] -3  5  3  6  7	-1	3
1 [3  -1  -3] 5  3  6  7	-3	3
1  3 [-1  -3  5] 3  6  7	-3	5
1  3  -1 [-3  5  3] 6  7	-3	5
1  3  -1  -3 [5  3  6] 7	3	6
1  3  -1  -3  5 [3  6  7]	3	7

Your task is to determine the maximum and minimum values in the sliding window at each position.

Input

The input consists of two lines. The first line contains two integers n and k which are the lengths of the array and the sliding window. There are n integers in the second line.

Output

There are two lines in the output. The first line gives the minimum values in the window at each position, from left to right, respectively. The second line gives the maximum values.

Sample Input

8 3 1 3 -1 -3 5 3 6 7 

Sample Output

-1 -3 -3 -3 3 3
3 3 5 5 6 7 

Source

POJ Monthly--2006.04.28, Ikki

题目大意：给一个长度为n的数组 ，和正整数k，然后依次从宽度（数的个数）为k的子组中找出最大值

 

//单调队列
#include<cstdio>
#include<queue>
using namespace std;
const int N=1e6;
int a[N+10],ma[N+10],mi[N+10];
int h[N+10],q[N+10];
int n,k;
void getmax()
{

    int i,head=1,tail=0;
    for( i=1;i<k;i++)
    {
        while(head<=tail&&h[tail]<a[i])//关键
            tail--;
        tail++;
        h[tail]=a[i];
        q[tail]=i;
    }
    for( i=k;i<=n;i++)
    {
        while(head<=tail&&h[tail]<a[i])
            tail--;
            tail++;
            h[tail]=a[i];
           q[tail]=i;
            while(q[head]<=i-k)head++;
            ma[i-k+1]=h[head];
    }
}
void getmin()
{
    int i,head=1,tail=0;
    for( i=1;i<k;i++)
    {
        while(head<=tail&&h[tail]>a[i])
            tail--;
        tail++;
        h[tail]=a[i];
        q[tail]=i;
    }
    for( i=k;i<=n;i++)
    {
         while(head<=tail&&h[tail]>a[i])
            tail--;
            tail++;
            h[tail]=a[i];
            q[tail]=i;
            while(q[head]<=i-k)head++;
            mi[i-k+1]=h[head];
    }
}
int main()
{

    while(~scanf("%d%d",&n,&k))
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        getmax();
        getmin();
   int aa=n-k+1;
    for(int i=1;i<aa;i++)
    printf("%d ",mi[i]);
    printf("%d\n",mi[aa]);
    for(int i=1;i<aa;i++)
    printf("%d ",ma[i]);
    printf("%d\n",ma[aa]);
    }

    return 0;
}

 

 

 

 优先队列的巧用    同理用堆也能写

 //优先队列 注意用G++提交会超时 用C++提交可以过
#include<iostream>
#include<cstdio>
#include<queue>
using namespace std;
const int N=1e6;
int a[N+10],ma[N+10],mi[N+10];//分别储存 原始数据 和 区间最大值 最小值
struct cmp1
{
    bool operator()(const int a1,const int a2)
    {
        return a[a1]>a[a2]; //越小越优先
    }
};
struct cmp2
{
    bool operator()(const int a1,const int a2)
    {
        return a[a1]<a[a2];
    }
};
int main()
{  int n,k;
    while(cin>>n>>k)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
    priority_queue<int,vector<int>,cmp1> q;
    priority_queue<int,vector<int>,cmp2> p;
    for(int i=1;i<=k;i++)
    {
        q.push(i);
        p.push(i);
    }
    int aa=0,ii=0;
    mi[++ii]=a[q.top()];
    ma[++aa]=a[p.top()];
    for(int i=k+1;i<=n;i++)
    {   int j=i-k;
        q.push(i);
        p.push(i);
        while(i-q.top()>=k)
        q.pop();  //pop()踢出当前最值
        while(i-p.top()>=k)
        p.pop();
    mi[++ii]=a[q.top()];
    ma[++aa]=a[p.top()];
    }
    for(int i=1;i<aa;i++)
        printf("%d ",mi[i]);
    printf("%d\n",mi[aa]);
    for(int i=1;i<aa;i++)
        printf("%d ",ma[i]);
    printf("%d\n",ma[aa]);
    }
    return 0;
}