HDU 3308 LCIS（线段树）

Problem Description

Given n integers.
You have two operations:
U A B: replace the Ath number by B. (index counting from 0)
Q A B: output the length of the longest consecutive increasing subsequence (LCIS) in [a, b].

 

Input

T in the first line, indicating the case number.
Each case starts with two integers n , m(0<n,m<=10 ⁵).
The next line has n integers(0<=val<=10 ⁵).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=10 ⁵)
OR
Q A B(0<=A<=B< n).

 

Output

For each Q, output the answer.

 

题目大意：给n个数，动态地修改某个数的值，或者查询一段区间的LCIS（最长连续上升子序列，坑爹的标题……）。

思路：线段树，每个点维护一个区间的从左边开始的最长的LCIS，从右边开始的最长的LCIS，这个区间最大的LCIS。然后随便搞搞就能过了……

 

代码（593MS）：

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <algorithm>

 5 using namespace std;

 6 

 7 const int MAXN = 100010 << 2;

 8 

 9 int lmax[MAXN], rmax[MAXN], mmax[MAXN];

10 int a[MAXN], n, m, T;

11 

12 void update(int x, int l, int r, int pos, int val) {

13     if(pos <= l && r <= pos) {

14         a[pos] = val;

15     } else {

16         int ll = x << 1, rr = ll | 1, mid = (l + r) >> 1;

17         if(pos <= mid) update(ll, l, mid, pos, val);

18         if(mid < pos) update(rr, mid + 1, r, pos, val);

19         if(a[mid] < a[mid + 1]) {

20             lmax[x] = lmax[ll] + (lmax[ll] == mid - l + 1) * lmax[rr];

21             rmax[x] = rmax[rr] + (rmax[rr] == r - mid) * rmax[ll];

22             mmax[x] = max(rmax[ll] + lmax[rr], max(mmax[ll], mmax[rr]));

23         } else {

24             lmax[x] = lmax[ll];

25             rmax[x] = rmax[rr];

26             mmax[x] = max(mmax[ll], mmax[rr]);

27         }

28     }

29 }

30 

31 int query(int x, int l, int r, int aa, int bb) {

32     if(aa <= l && r <= bb) {

33         return mmax[x];

34     } else {

35         int ll = x << 1, rr = ll | 1, mid = (l + r) >> 1;

36         int ans = 0;

37         if(aa <= mid) ans = max(ans, query(ll, l, mid, aa, bb));

38         if(mid < bb) ans = max(ans, query(rr, mid + 1, r, aa, bb));

39         if(a[mid] < a[mid + 1]) ans = max(ans, min(rmax[ll], mid - aa + 1) + min(lmax[rr], bb - mid));

40         return ans;

41     }

42 }

43 

44 void build(int x, int l, int r) {

45     if(l == r) {

46         lmax[x] = rmax[x] = mmax[x] = 1;

47     } else {

48         int ll = x << 1, rr = ll | 1, mid = (l + r) >> 1;

49         build(ll, l, mid);

50         build(rr, mid + 1, r);

51         if(a[mid] < a[mid + 1]) {

52             lmax[x] = lmax[ll] + (lmax[ll] == mid - l + 1) * lmax[rr];

53             rmax[x] = rmax[rr] + (rmax[rr] == r - mid) * rmax[ll];

54             mmax[x] = max(rmax[ll] + lmax[rr], max(mmax[ll], mmax[rr]));

55         } else {

56             lmax[x] = lmax[ll];

57             rmax[x] = rmax[rr];

58             mmax[x] = max(mmax[ll], mmax[rr]);

59         }

60     }

61 }

62 

63 int main() {

64     scanf("%d", &T);

65     while(T--) {

66         scanf("%d%d", &n, &m);

67         for(int i = 0; i < n; ++i) scanf("%d", &a[i]);

68         build(1, 0, n - 1);

69         while(m--) {

70             char c;

71             int a, b;

72             scanf(" %c%d%d", &c, &a, &b);

73             if(c == 'Q') printf("%d\n", query(1, 0, n - 1, a, b));

74             else update(1, 0, n - 1, a, b);

75         }

76     }

77 }

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