hdu 3530 Subsequence

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=3530

Subsequence

Description

There is a sequence of integers. Your task is to find the longest subsequence that satisfies the following condition: the difference between the maximum element and the minimum element of the subsequence is no smaller than m and no larger than k.

Input

There are multiple test cases.
For each test case, the first line has three integers, n, m and k. n is the length of the sequence and is in the range $[1, 100000]$. m and k are in the range $[0, 1000000]$. The second line has n integers, which are all in the range $[0, 1000000]$.
Proceed to the end of file.

Output

For each test case, print the length of the subsequence on a single line.

Sample Input

5 0 0
1 1 1 1 1
5 0 3
1 2 3 4 5

Sample Output

5
4

RMQ线段树。。

 1 #include<algorithm>

 2 #include<iostream>

 3 #include<cstdlib>

 4 #include<cstring>

 5 #include<cstdio>

 6 #include<vector>

 7 #include<map>

 8 #include<set>

 9 using std::set;

10 using std::map;

11 using std::min;

12 using std::max;

13 using std::cin;

14 using std::cout;

15 using std::endl;

16 using std::find;

17 using std::sort;

18 using std::pair;

19 using std::vector;

20 #define sz(c) (int)(c).size()

21 #define all(c) (c).begin(), (c).end()

22 #define iter(c) decltype((c).begin())

23 #define cls(arr,val) memset(arr,val,sizeof(arr))

24 #define cpresent(c, e) (find(all(c), (e)) != (c).end())

25 #define rep(i, n) for (int i = 1; i <= (int)(n); i++)

26 #define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)

27 #define pb(e) push_back(e)

28 #define mp(a, b) make_pair(a, b)

29 #define mid ((l+r)>>1)

30 #define lc (root<<1)

31 #define rc (root<<1|1)

32 const int Max_N = 100010;

33 const int INF = 0x3f3f3f3f;

34 typedef unsigned long long ull;

35 int n, m, k, tmin, tmax;

36 struct SegTree {

37     struct Node { int max, min; }seg[Max_N << 2];

38     inline void built(int root, int l, int r) {

39         if (l == r) {

40             scanf("%d", &seg[root].max), seg[root].min = seg[root].max;

41             return;

42         }

43         built(lc, l, mid);

44         built(rc, mid + 1, r);

45         seg[root].max = max(seg[lc].max, seg[rc].max);

46         seg[root].min = min(seg[lc].min, seg[rc].min);

47     }

48     inline void query(int root, int l, int r, int x, int y) {

49         if (x > r || y < l) return;

50         if (x <= l && y >= r) {

51             tmin = min(tmin, seg[root].min);

52             tmax = max(tmax, seg[root].max);

53             return;

54         }

55         query(lc, l, mid, x, y);

56         query(rc, mid + 1, r, x, y);

57     }

58     inline int query(int l, int r) {

59         tmin = INF, tmax = -INF;

60         query(1, 1, n, l, r);

61         return tmax - tmin;

62     }

63 }seg;

64 int main() {

65 #ifdef LOCAL

66     freopen("in.txt", "r", stdin);

67     freopen("out.txt", "w+", stdout);

68 #endif

69     int l, r, ans;

70     while (~scanf("%d %d %d", &n, &m, &k)) {

71         l = 1, ans = 0;

72         seg.built(1, 1, n);

73         rep(i, n) {

74             r = i;

75             if (l > r) continue;

76             while (seg.query(l, r) > k) l++;

77             if (seg.query(l, r) >= m && seg.query(l, r) <= k) ans = max(ans, r - l + 1);

78         }

79         printf("%d\n", ans);

80     }

81     return 0;

82 }

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