单调队列问题「建议收藏」

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 1000005;
int a[maxn];
int Min[maxn];//记录每个窗口最小答案
int Max[maxn];//记录每个窗口最大答案
int que[maxn];//数组模拟单调队列
int pos[maxn];//记录位置，用来更新

int n, k;

void get_min() {
  
  //维护单调递增队列，队头为最小
    int head = 0, rear = 0;
    for(int i = 1; i < k; i ++) {
  
  //先将前k-1个放入队列
        while(rear > head && que[rear - 1] > a[i]) rear --;
        que[rear] = a[i];
        pos[rear] = i;
        rear ++;
    }
    for(int i = k; i <= n; i ++) {
        while(rear > head && que[rear - 1] > a[i]) rear --;
        que[rear] = a[i];
        pos[rear] = i;
        rear ++;
        while(pos[head] < i - k + 1) head ++;//因为是滑窗，所以要及时更新
        Min[i] = que[head];
    }
}

void get_max() {
  
  //维护单调递减队列，队头为最大
    int head = 0, rear = 0;
    for(int i = 1; i < k; i ++) {
  
  //先将前k-1个放入队列
        while(rear > head && que[rear - 1] < a[i]) rear --;
        que[rear] = a[i];
        pos[rear] = i;
        rear ++;
    }
    for(int i = k; i <= n; i ++) {
        while(rear > head && que[rear - 1] < a[i]) rear --;
        que[rear] = a[i];
        pos[rear] = i;
        rear ++;
        while(pos[head] < i - k + 1) head ++;//因为是滑窗，所以要及时更新
        Max[i] = que[head];
    }
}

int main() {
    while(scanf("%d %d", &n, &k) != EOF) {
        for(int i = 1; i <= n; i ++) {
            scanf("%d", &a[i]);
        }
        get_min();
        get_max();
        for(int i = k; i < n; i ++) {
            printf("%d ", Min[i]);
        }
        printf("%d\n", Min[n]);
        for(int i = k; i < n; i ++) {
            printf("%d ", Max[i]);
        }
        printf("%d\n", Max[n]);
    }
    return 0;
}

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <fstream>
#include <sstream>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 100005;
int n, m, k;
int a[maxn];
int q1[maxn], q2[maxn];//单调队列记录最大和最小的位置
int head1, rear1;
int head2, rear2;

int main() {
    while(scanf("%d %d %d", &n, &m, &k) != EOF) {
        head1 = rear1 = 0;
        head2 = rear2 = 0;
        int pre = 1;//记录可行区间的最左,i可以动态表示可行区间的最右
        int ans = 0;
        for(int i = 1;  i <= n; i ++) {
            scanf("%d", &a[i]);
            while(rear1 > head1 && a[q1[rear1 - 1]] < a[i]) rear1 --;
            while(rear2 > head2 && a[q2[rear2 - 1]] > a[i]) rear2 --;
            q1[rear1 ++] = i;
            q2[rear2 ++] = i;
            while(rear1 > head1 && rear2 > head2 && a[q1[head1]] - a[q2[head2]] > k) {
  
  //更新可行区间的最左
                if(q1[head1] > q2[head2]) pre = q2[head2 ++] + 1;
                else pre = q1[head1 ++] + 1;
            }
            if(rear1 > head1 && rear2 > head2 && a[q1[head1]] - a[q2[head2]] >= m) {
  
  //此区间的最大最小满足在m,k之间，所以更新答案
                ans = max(ans, i - pre + 1);//更新答案
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}