双端队列 HDOJ 3530 Subsequence

题目传送门

题意：问最长子序列，满足区间最大值 - 最小值在[m, k]之间

分析：用双端队列维护最大值和最小值，保存的是位置。当满足条件时，更新最大值。

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/25 星期五 08:50:32
* File Name     :A_deque.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int a[N];

int main(void)    {
    int n, m, k;
    while (scanf ("%d%d%d", &n, &m, &k) == 3)   {
        deque<int> Q1, Q2;
        int ans = 0, l = 1;
        for (int i=1; i<=n; ++i) {
            scanf ("%d", &a[i]);
            while (!Q1.empty () && a[Q1.back ()] <= a[i])    Q1.pop_back ();
            Q1.push_back (i);
            while (!Q2.empty () && a[Q2.back ()] >= a[i])    Q2.pop_back ();
            Q2.push_back (i);
            while (!Q1.empty () && !Q2.empty () && a[Q1.front ()] - a[Q2.front ()] > k)   {
                if (Q1.front () < Q2.front ())  {
                    l = Q1.front () + 1;    Q1.pop_front ();
                }
                else    {
                    l = Q2.front () + 1;    Q2.pop_front ();
                }
            }
            if (!Q1.empty () && !Q2.empty () && a[Q1.front ()] - a[Q2.front ()] >= m)   {
                if (ans < i - l + 1)    ans = i - l + 1;
            }
        }
        printf ("%d\n", ans);
    }

    return 0;
}

数组模拟：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/22 星期二 15:49:25
* File Name     :A.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
int dq1[N], dq2[N], a[N];

int main(void)    {
    int n, m, k;
    while (scanf ("%d%d%d", &n, &m, &k) == 3)   {
        for (int i=1; i<=n; ++i)    {
            scanf ("%d", &a[i]);
        }
        int f1 = 0, f2 = 0, b1 = 0, b2 = 0, l = 1;
        int ans = 0;
        for (int i=1; i<=n; ++i)    {
            while (f1 < b1 && a[dq1[b1-1]] <= a[i])  b1--;
            dq1[b1++] = i;
            while (f2 < b2 && a[dq2[b2-1]] >= a[i])  b2--;
            dq2[b2++] = i;
            while (f1 < b1 && f2 < b2 && a[dq1[f1]] - a[dq2[f2]] > k) {
                if (dq1[f1] < dq2[f2])  {
                    l = dq1[f1++] + 1;
                }
                else    l = dq2[f2++] + 1;
            }
            if (f1 < b1 && f2 < b2 && a[dq1[f1]] - a[dq2[f2]] >= m)   {
                ans = max (ans, i - l + 1);
            }
        }
        printf ("%d\n", ans);
    }

    return 0;
}