[hdu3308]线段树

题意：单点更新，区间LCIS(最长连续递增序列)查询。具备区间合并维护的性质，不用线段树用什么~

#pragma comment(linker, "/STACK:10240000,10240000")

#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;
typedef vector<int> vi;

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 = 1e5 + 7;
const int maxm = 1e5 + 7;
const int maxv = 1e7 + 7;
const int MD = 1e9 +7;
const int INF = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-10;

int A[maxn];

struct SegTree {
    struct Node {
        int suflen, prelen, len;
    } tree[maxn << 2];

    Node merge(Node a, Node b, int l, int m, int r) {
        Node c;
        int leftLen = m - l + 1, rightLen = r - m;
        c.prelen = a.prelen;
        if (c.prelen == leftLen && A[m] < A[m + 1]) c.prelen += b.prelen;
        c.suflen = b.suflen;
        if (c.suflen == rightLen && A[m] < A[m + 1]) c.suflen += a.suflen;
        c.len = a.len;
        c.len = max(c.len, b.len);
        if (A[m] < A[m + 1])c.len = max(c.len, a.suflen + b.prelen);
        return c;
    }
    void build(int l, int r, int rt) {
        if (l == r) {
            int x;
            scanf("%d", &x);
            A[l] = x;
            tree[rt].len = tree[rt].prelen =  tree[rt].suflen = 1;
            return ;
        }
        define_m;
        build(lson);
        build(rson);
        tree[rt] = merge(tree[rt << 1], tree[rt << 1 | 1], l, m, r);
    }
    void update(int p, int x, int l, int r, int rt) {
        if (l == r) {
            tree[rt].len = tree[rt].prelen =  tree[rt].suflen = 1;
            A[l] = x;
            return ;
        }
        define_m;
        if (p <= m) update(p, x, lson);
        else update(p, x, rson);
        tree[rt] = merge(tree[rt << 1], tree[rt << 1 | 1], l, m, r);
    }
    Node query(int L, int R, int l, int r, int rt) {
        if (L <= l && r <= R) return tree[rt];
        define_m;
        if (R <= m) return query(L, R, lson);
        if (L > m) return query(L, R, rson);
        return merge(query(L, m, lson), query(m + 1, R, rson), L, m, R);
    }
};

SegTree st;

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    cin >> T;
    while (T--) {
        int n, m;
        cin >> n >> m;
        st.build(1, n, 1);
        for (int i = 0, u, v; i < m; i++) {
            char s[3];
            scanf("%s%d%d", s, &u, &v);
            if (s[0] == 'U') {
                st.update(++u, v, 1, n, 1);
            }
            else {
                printf("%d
", st.query(++u, ++v, 1, n, 1).len);
            }
        }
    }
    return 0;
}