POJ 2823 Sliding Window(单调队列)

单调队列典型题

An array of size n ≤ 10 ⁶ is given to you. There is a sliding window of sizek which is moving from the very left of the array to the very right. You can only see thek 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.

#include<cstdio>  #include<cstring>  #include<cmath>  #include<cstdlib>  #include<iostream>  #include<algorithm>  #include<vector>  #include<map>  #include<queue>  #include<stack> #include<string>#include<map> #include<set>#define eps 1e-6 #define LL long long  using namespace std;  const int maxn = 2000000;//const int INF = 0x3f3f3f3f;int n, k;int a[maxn];//单调队列int qmin[maxn], vmin[maxn], hmin = 1, tmin = 0; void Min(int a, int i) {  //第i个元素a入队 while(hmin<=tmin && vmin[hmin] <= i-k) hmin++;  //超范围队首出队 //while(hmin<=tmin && qmin[tmin]>=a) tmin--; //不符合要求队尾出列 int l = hmin, r = tmin;while(l <= r) {int m = l+(r-l)/2;if(qmin[m] >= a) r = m - 1;else l = m + 1; }tmin = ++r;qmin[tmin] = a;vmin[tmin] = i;}int qmax[maxn], vmax[maxn], hmax = 1, tmax = 0; void Max(int a, int i) {  //第i个元素a入队 while(hmax<=tmax && vmax[hmax] <= i-k) hmax++;  //超范围队首出队 //while(hmax<=tmax && qmax[tmax]<=a) tmax--; //不符合要求队尾出列 int l = hmax, r = tmax;while(l <= r) {int m = l+(r-l)/2;if(qmax[m] <= a) r = m - 1;else l = m + 1; }tmax = ++r;qmax[tmax] = a;vmax[tmax] = i;}int ansMax[maxn], ansMin[maxn];int main() {//freopen("input.txt", "r", stdin);while(scanf("%d%d", &n, &k) == 2) {hmin = 1, tmin = 0;hmax = 1, tmax = 0;for(int i = 1; i <= n; i++) scanf("%d", &a[i]);for(int i = 1; i < k; i++) {Min(a[i], i);Max(a[i], i);}for(int i = k; i <= n; i++) {Min(a[i], i);ansMin[i-k] = qmin[hmin];Max(a[i], i);ansMax[i-k] = qmax[hmax];}for(int i = 0; i <= n-k; i++) if(i != n-k) printf("%d ", ansMin[i]);else printf("%d\n", ansMin[i]);for(int i = 0; i <= n-k; i++) if(i != n-k) printf("%d ", ansMax[i]);else printf("%d\n", ansMax[i]);}return 0;}