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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<=105).
The next line has n integers(0<=val<=105).
The next m lines each has an operation:
U A B(0<=A,n , 0<=B=105)
OR
Q A B(0<=A<=B< n).

Output

For each Q, output the answer.

Sample Input

1
10 10
7 7 3 3 5 9 9 8 1 8
Q 6 6
U 3 4
Q 0 1
Q 0 5
Q 4 7
Q 3 5
Q 0 2
Q 4 6
U 6 10
Q 0 9

Sample Output

1
1
4
2
3
1
2
5

题意

给定n个整数。
你有两种操作：

  • U A B：用B替换第A个数。（编号从0开始）
  • Q A B：输出[a，b]中最长连续严格递增子序列（LCIS）的长度。

题解

  • 首先序列严格递增，7,7这样的序列不算递增
  • 其次本题是单点更新,所以不用PushDown,只用PushUp就行.
  • 线段树维护信息:
    mx：最长上升子序列的长度)
    lm：包含区间左端点的LIS长度
    rm：包含区间右端点的LIS长度
  • PushUp的时候
  • lm、rm直接从子节点获得即可
  • mx显然可以从左右两个子区间的最大值得到。
    还有一种可能是左右区间各取一部分，此时必须满足左区间的右端点值严格小于右区间的左端点值。
  • 疑惑:为何维护的lm、rm是左端点、右端点的LIS，为何不是左右孩子节点的LIS？

AC-Code

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <fstream>
#include <utility>
using namespace std;
typedef long long ll;
typedef pair<int, int> Pii;
#define

const int INF = 0x7fffffff;
const int MOD = 1e4;
const int MAXN = 1e5 + 2;

struct node {
  int left;
  int right;
  int mx;
  int lm;
  int rm;
}tree[MAXN << 2];
int n, m;
int a[MAXN];
void PushUp(int rt) {
  tree[rt].lm = tree[rt << 1].lm;
  tree[rt].rm = tree[rt << 1 | 1].rm;
  tree[rt].mx = max(tree[rt << 1].mx, tree[rt << 1 | 1].mx);
  int mid = (tree[rt].left + tree[rt].right) >> 1;
  if (a[mid] < a[mid + 1])
  {
    if (tree[rt << 1].lm == mid - tree[rt].left + 1) tree[rt].lm += tree[rt << 1 | 1].lm;
    if (tree[rt << 1 | 1].rm == tree[rt].right - mid) tree[rt].rm += tree[rt << 1].rm;
    tree[rt].mx = max(tree[rt].mx, tree[rt << 1].rm + tree[rt << 1 | 1].lm);
  }
}

void build(int rt, int l, int r) {
  tree[rt].left = l;
  tree[rt].right = r;
  if (l == r) {
    tree[rt].mx = tree[rt].lm = tree[rt].rm = 1;
    return;
  }
  int mid = (l + r) >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
  PushUp(rt);
}
void updata(int rt, int val, int p) {
  if (tree[rt].left == tree[rt].right) {
    a[p] = val;
    return;
  }
  int mid = (tree[rt].left + tree[rt].right) >> 1;
  if (p <= mid)  updata(rt << 1, val, p);
  else  updata(rt << 1 | 1, val, p);
  PushUp(rt);
}
int Query(int rt, int l, int r) {
  if (l <= tree[rt].left && tree[rt].right <= r) {
    return tree[rt].mx;
  }
  int mid = (tree[rt].left + tree[rt].right) >> 1;
  int s = 0;
  if (l <= mid)  s = max(s, Query(rt << 1, l, r));
  if (r > mid)  s = max(s, Query(rt << 1 | 1, l, r));
  if (a[mid] < a[mid + 1])
    s = max(s, min(tree[rt << 1].rm, mid - l + 1) + min(tree[rt << 1 | 1].lm, r - mid));
  return s;
}

int main() {
  ios;
  int T;
  cin >> T;
  while (T--) {
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
      cin >> a[i];
    build(1, 1, n);
    while (m--) {
      char str[3];
      int x, y;
      cin >> str >> x >> y;
      if (str[0] == 'Q')
        cout << Query(1, x + 1, y + 1) << endl;
      else
        updata(1, y, x + 1);
    }

  }
  return 0;
}