HDU 3530 单调队列

最新推荐文章于 2019-07-25 09:06:47 发布

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最新推荐文章于 2019-07-25 09:06:47 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/CJLHY/p/9173520.html

本文介绍了一种使用单调队列解决特定区间最值问题的方法。通过维护递增和递减排列的队列，可以在O(n)的时间复杂度内找到满足条件的最大区间长度。

题目大意：给你n个数，让你问你最长的满足要求的区间有多长，区间要求：MAX - MIN >= m && MAX - MIN <= k

思路：单调队列维护递增和递减，在加入数值的过程中更新答案。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> >

using namespace std;

const int N = 1e5 + 7;
const int M = 10 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int n, m, k, head1, head2, rear1, rear2;
int sk1[N], sk2[N], a[N];

void init() {
    head1 = head2 = 0;
    rear1 = rear2 = 0;
}

int main() {
    while(scanf("%d%d%d", &n, &m, &k) != EOF) {

        init();

        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
        }

        int ans = 0, now = 1;
        for(int i = 1; i <= n; i++) {
            while(head1 < rear1 && a[sk1[rear1 - 1]] < a[i]) rear1--;
            while(head2 < rear2 && a[sk2[rear2 - 1]] > a[i]) rear2--;
            sk1[rear1++] = i;
            sk2[rear2++] = i;

            while(head1 < rear1 && head2 < rear2 && a[sk1[head1]] - a[sk2[head2]] > k) {

                if(sk1[head1] < sk2[head2]) now = sk1[head1++] + 1;
                else now = sk2[head2++] + 1;
            }

            if(head2 < rear2 && head2 < rear2 && a[sk1[head1]] - a[sk2[head2]] >= m) {

                ans = max(ans, i - now + 1);
            }
        }

        printf("%d\n", ans);
    }
    return 0;
}
/*
*/