POJ 2823 Sliding Window（单调队列）

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 positionMinimum valueMaximum value[1  3  -1] -3  5  3  6  7 -13 1 [3  -1  -3] 5  3  6  7 -33 1  3 [-1  -3  5] 3  6  7 -35 1  3  -1 [-3  5  3] 6  7 -35 1  3  -1  -3 [5  3  6] 7 36 1  3  -1  -3  5 [3  6  7]37

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 31 3 -1 -3 5 3 6 7

Sample Output

-1 -3 -3 -3 3 33 3 5 5 6 7

一.总结

首先对于这道题来说，一开始的时候其实考虑了解法，但是还是没有深入下去对解法进行设计并实现这个解法，这道题运用了stl的优先队列，并且运用了结构题做元素，很经典，需要重写cmp并重载（）

#include <iostream>#include <queue>#include <cstdio>#include <cstring>#include <cmath>using namespace std;const int maxn=1000010;struct Node{    int data;    int p;}nodes[maxn];struct cmp1{    bool operator() (const Node a,const Node b) const    {        return a.data>b.data;    }};struct cmp2{    bool operator() (const Node a,const Node b) const    {        return a.data<b.data;    }};int main(){    //freopen("/Users/zhangjiatao/Documents/暑期训练/input.txt","r",stdin);    int n,k;    while(scanf("%d%d",&n,&k)!=EOF)    {        int a1[maxn],a2[maxn],counter=0;        priority_queue <Node,vector<Node>,cmp1> p1;        priority_queue <Node,vector<Node>,cmp2> p2;        for(int i=1;i<=n;i++)        {            cin>>nodes[i].data;            nodes[i].p=i;        }        for(int i=1;i<=k;i++) {p1.push(nodes[i]);p2.push(nodes[i]);}        a1[++counter]=p1.top().data;        a2[counter]=p2.top().data;        for(int i=k+1;i<=n;i++)        {            p1.push(nodes[i]);            p2.push(nodes[i]);            while(1)            {                if(i-p1.top().p<k)                {                    a1[++counter]=p1.top().data;                    break;                }                else p1.pop();            }            while(1)            {                if(i-p2.top().p<k)                {                    a2[counter]=p2.top().data;                    break;                }                else p2.pop();            }        }        for(int i=1;i<=counter;i++) printf("%d ",a1[i]);        cout<<endl;        for(int i=1;i<=counter;i++) printf("%d ",a2[i]);        cout<<endl;    }    return 0;}