soj 3139 Sliding Window(单调队列)

@(K ACMer)
题意
维护固定长度子区间的最大最小值.
分析
最经典的单调队列的应用,其实究其根本是用了记忆化和剪枝的思想,使整个效率很高.剪枝的思想,就是对于下标小于当前数,且值大于当前数的数,永远没有用(维护最小值的时候).

code

#include <iostream>
#include <cstdio>
#include <cstring>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <string>
#include <queue>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 1e6 + 40;
int n, k, a[maxn], dq[maxn * 2], dqq[maxn * 2], ansmin[maxn], ansmax[maxn];

int main(void) {
    while (~scanf("%d%d", &n, &k)) {
        for (int i = 0; i < n; i++)
            scanf("%d", &a[i]);
        int s = 0, t = -1, ss = 0, tt = 0;
        for (int i = 0; i < n; i++) {
            while (t != -1 && a[i] < a[dq[t]] && t >= s) t--;
            while (tt > ss && a[i] > a[dqq[tt - 1]]) tt--;
            dq[++t] = i;
            dqq[tt++] = i;
            if (i - k >= dq[s]) s++;
            if (i - k >= dqq[ss]) ss++;
            if (i >= k - 1) ansmin[i] = dq[s], ansmax[i] = dqq[ss];
        }

        for (int i = k - 1; i < n; i++) {
            if (i != k - 1) printf(" ");
            printf("%d", a[ansmin[i]]);
        }
        puts("");

        for (int i = k - 1; i < n; i++) {
            if (i != k - 1) printf(" ");
            printf("%d", a[ansmax[i]]);
        }
        puts("");
    }
    return 0;
}