POJ 2823 单调队列入门水题

最最基础的单调队列题目。一个单增一个单减。还是可以借此好好理解一下单调队列的。

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

#define maxx 1000005

int num[maxx], inque[maxx], dque[maxx], maxn[maxx], minn[maxx];
int pre1, pre2, lst1, lst2;
int n, k;

int main() {
    while(~scanf("%d%d", &n, &k)) {
        pre1 = 0, pre2 = 0, lst1 = 0, lst2 = 0;
        int cnt = 0;
        for (int i=0; i<n; ++i) {
            scanf("%d", &num[i]);
            // 单增
            while(pre1 < lst1 && num[inque[lst1-1]] > num[i])
                lst1--;
            while(pre2 < lst2 && num[dque[lst2-1]] < num[i])
                lst2--;
            inque[lst1++] = i;
            dque[lst2++] = i;
            while(pre1 < lst1 && i - inque[pre1] + 1 > k)
                pre1++;
            while(pre2 < lst2 && i - dque[pre2] + 1 > k)
                pre2++;
            if (i + 1 >= k) {
                maxn[cnt] = num[inque[pre1]];
                minn[cnt] = num[dque[pre2]];
                cnt++;
            }
        }
        for (int i=0; i<cnt-1; ++i) {
            printf("%d ", maxn[i]);
        }
        printf("%d
", maxn[cnt-1]);
        for (int i=0; i<cnt-1; ++i) {
            printf("%d ", minn[i]);
        }
        printf("%d
", minn[cnt-1]);
    }
    return 0;
}