[hdu3530]单调队列

题意：求满足最大元素与最小元素之差在一定范围的连续区间的最大长度。单调队列经典应用。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define Rep(a, b) for(int a = 0; a < b; a++)
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}

typedef double db;
typedef long long LL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef multiset<int>::iterator msii;
typedef set<int> si;
typedef set<int>::iterator sii;

const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1};
const int maxn = 1e6 + 7;
const int maxm = 1e5 + 7;
const int MD = 1e9 +7;
const int INF = 1e9 + 7;

template<class T> struct MonotoneQueue{
    deque<T> Q;
    MonotoneQueue<T>() { Q.clear(); }
    void clear() { Q.clear(); }
    bool empty() { return Q.empty(); }
    void add_back(T x) { while (!Q.empty() && !(Q.back() < x)) Q.pop_back(); Q.push_back(x); }
    void pop_front() { Q.pop_front(); }
    T front() { return Q.front(); }
};

struct Pair1 {
    int val, pos;
    bool operator < (const Pair1 &a) const {
        return val < a.val;
    }
    constructInt2(Pair1, val, pos);
};
struct Pair2 {
    int val, pos;
    bool operator < (const Pair2 &a) const {
        return val > a.val;
    }
    constructInt2(Pair2, val, pos);
};
MonotoneQueue<Pair1> UQ;
MonotoneQueue<Pair2> DQ;

int main() {
    //freopen("in.txt", "r", stdin);
    int n, m, k;
    while (cin >> n >> m >> k) {
        UQ.clear(); DQ.clear();
        int L = 0, ans = 0;
        for (int i = 0, x; i < n; i++) {
            scanf("%d", &x);
            UQ.add_back(Pair1(x, i));
            DQ.add_back(Pair2(x, i));
            while (DQ.front().val - UQ.front().val > k) {
                L++;
                if (UQ.front().pos < L) UQ.pop_front();
                if (DQ.front().pos < L) DQ.pop_front();
            }
            if (DQ.front().val - UQ.front().val >= m) ans = max(ans, i - L + 1);
        }
        cout << ans << endl;
    }
    return 0;
}