[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。
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <map>
 7 #include <queue>
 8 #include <deque>
 9 #include <cmath>
10 #include <vector>
11 #include <ctime>
12 #include <cctype>
13 #include <set>
14 
15 using namespace std;
16 
17 #define mem0(a) memset(a, 0, sizeof(a))
18 #define lson l, m, rt << 1
19 #define rson m + 1, r, rt << 1 | 1
20 #define define_m int m = (l + r) >> 1
21 #define Rep(a, b) for(int a = 0; a < b; a++)
22 #define lowbit(x) ((x) & (-(x)))
23 #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) {}
24 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
25 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
26 
27 typedef double db;
28 typedef long long LL;
29 typedef pair<int, int> pii;
30 typedef multiset<int> msi;
31 typedef multiset<int>::iterator msii;
32 typedef set<int> si;
33 typedef set<int>::iterator sii;
34 
35 const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
36 const int dy[8] = {0, -1, 0, 1, -1, 1, 1, -1};
37 const int maxn = 1e6 + 7;
38 const int maxm = 1e5 + 7;
39 const int MD = 1e9 +7;
40 const int INF = 1e9 + 7;
41 
42 template<class T> struct MonotoneQueue{
43     deque<T> Q;
44     MonotoneQueue<T>() { Q.clear(); }
45     void clear() { Q.clear(); }
46     bool empty() { return Q.empty(); }
47     void add_back(T x) { while (!Q.empty() && !(Q.back() < x)) Q.pop_back(); Q.push_back(x); }
48     void pop_front() { Q.pop_front(); }
49     T front() { return Q.front(); }
50 };
51 
52 struct Pair1 {
53     int val, pos;
54     bool operator < (const Pair1 &a) const {
55         return val < a.val;
56     }
57     constructInt2(Pair1, val, pos);
58 };
59 struct Pair2 {
60     int val, pos;
61     bool operator < (const Pair2 &a) const {
62         return val > a.val;
63     }
64     constructInt2(Pair2, val, pos);
65 };
66 MonotoneQueue<Pair1> UQ;
67 MonotoneQueue<Pair2> DQ;
68 
69 int main() {
70     //freopen("in.txt", "r", stdin);
71     int n, m, k;
72     while (cin >> n >> m >> k) {
73         UQ.clear(); DQ.clear();
74         int L = 0, ans = 0;
75         for (int i = 0, x; i < n; i++) {
76             scanf("%d", &x);
77             UQ.add_back(Pair1(x, i));
78             DQ.add_back(Pair2(x, i));
79             while (DQ.front().val - UQ.front().val > k) {
80                 L++;
81                 if (UQ.front().pos < L) UQ.pop_front();
82                 if (DQ.front().pos < L) DQ.pop_front();
83             }
84             if (DQ.front().val - UQ.front().val >= m) ans = max(ans, i - L + 1);
85         }
86         cout << ans << endl;
87     }
88     return 0;
89 }
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