POJ 2823 Sliding Window（滑动窗口问题__优先队列||单调队列）
Description
An array of size
n ≤ 10
6 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.
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
<pre name="code" class="html">#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int maxn=1e6+100;
int a[maxn],ans1[maxn],ans2[maxn];
struct cmp1
{
bool operator ()(int x,int y)
{
return a[x]>a[y];
}
};
struct cmp2
{
bool operator ()(int x,int y)
{
return a[x]<a[y];
}
};
priority_queue<int,vector<int>,cmp1>qq1;
priority_queue<int,vector<int>,cmp2>qq2;
int main()
{
int n,k,i,j;
while(scanf("%d%d",&n,&k)!=EOF) {
int kk=0;
for(i=1;i<=n;i++) scanf("%d",&a[i]);
for(i=1;i<=k;i++) {
qq1.push(i);
qq2.push(i);
}
ans1[++kk]=a[qq1.top()];
ans2[kk]=a[qq2.top()];
for(i=k+1;i<=n;i++) {
qq1.push(i);
qq2.push(i);
while(i-qq1.top()>=k) qq1.pop(); //很巧妙的方法，一开始想的是怎么删去超过边界的元素。
ans1[++kk]=a[qq1.top()]; //换个思路，如何当前的最小值不在范围内，就把它删去！！！
while(i-qq2.top()>=k) qq2.pop(); //删除已经移除届的数 注意此处理解
ans2[kk]=a[qq2.top()]; //Qmin.top()为元素最小值的下标，
} //若i与最小值下标间相差大于等于K，即最小值元素已删除，此时应Qmin.pop();
for(i=1;i<=kk;i++){
if(i>1) printf(" %d",ans1[i]);
else printf("%d",ans1[i]);
}
printf("\n");
for(i=1;i<=kk;i++){
if(i>1) printf(" %d",ans2[i]);
else printf("%d",ans2[i]);
}
printf("\n");
// while(!qq1.empty()) qq1.pop();
// while(!qq2.empty()) qq2.pop();
}
return 0;
}