HDU 3308 LCIS

区间合并。给出一些数，记为a[i]，两种操作。U x y表示把a[x] 的值改为 y。Q x y表示求xy间的最长连续上升序列（LCIS）。这里的x y都是从0开始的。

其实吧，就是比较当前区间的左右子区间能否相连，也就是说要看a[k] 与 a[k + 1]的大小关系（k表示区间中点）。其他操作跟“正常”题一样。

 

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<vector>

#include<queue>

#include<cmath>

///LOOP

#define REP(i, n) for(int i = 0; i < n; i++)

#define FF(i, a, b) for(int i = a; i < b; i++)

#define FFF(i, a, b) for(int i = a; i <= b; i++)

#define FD(i, a, b) for(int i = a - 1; i >= b; i--)

#define FDD(i, a, b) for(int i = a; i >= b; i--)

///INPUT

#define RI(n) scanf("%d", &n)

#define RII(n, m) scanf("%d%d", &n, &m)

#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)

#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)

#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)

#define RFI(n) scanf("%lf", &n)

#define RFII(n, m) scanf("%lf%lf", &n, &m)

#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)

#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)

#define RS(s) scanf("%s", s)

///OUTPUT

#define PN printf("\n")

#define PI(n) printf("%d\n", n)

#define PIS(n) printf("%d ", n)

#define PS(s) printf("%s\n", s)

#define PSS(s) printf("%s ", n)

#define PC(n) printf("Case %d: ", n)

///OTHER

#define PB(x) push_back(x)

#define CLR(a, b) memset(a, b, sizeof(a))

#define CPY(a, b) memcpy(a, b, sizeof(b))

#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}

#define lson l, m, rt << 1

#define rson m + 1, r, rt << 1 | 1



using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int MOD = 9901;

const int INFI = 1e9 * 2;

const LL LINFI = 1e17;

const double eps = 1e-6;

const double pi = acos(-1.0);

const int N = 111111;

const int M = 22;

const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};



int a[N], msum[N << 2], lsum[N << 2], rsum[N << 2], num, n;



void pushup(int rt, int k, int m)

{

    lsum[rt] = lsum[rt << 1];

    rsum[rt] = rsum[rt << 1 | 1];

    msum[rt] = max(msum[rt << 1], msum[rt << 1 | 1]);

    if(a[k] < a[k + 1])

    {

        if(lsum[rt] == (m - (m >> 1)))lsum[rt] += lsum[rt << 1 | 1];

        if(rsum[rt] == (m >> 1))rsum[rt] += rsum[rt << 1];

        msum[rt] = max(msum[rt], lsum[rt << 1 | 1] + rsum[rt << 1]);

    }

}



void build(int l, int r, int rt)

{

    if(l == r)

    {

        RI(a[num++]);

        msum[rt] = rsum[rt] = lsum[rt] = 1;

        return;

    }

    int m = (l + r) >> 1;

    build(lson);

    build(rson);

    pushup(rt, m, r - l + 1);

}



void update(int p, int x, int l, int r, int rt)

{

    if(l == r)

    {

        a[p] = x;

        return;

    }

    int m = (l + r) >> 1;

    if(p <= m)update(p, x, lson);

    else update(p, x, rson);

    pushup(rt, m, r - l + 1);

}



int query(int L, int R, int l, int r, int rt)

{

    if(L <= l && r <= R)return msum[rt];

    int m = (l + r) >> 1, ans = 0;

    if(L <= m)ans = max(ans, query(L, R, lson));

    if(R > m)ans = max(ans, query(L, R, rson));

    if(a[m] < a[m + 1] && L <= m && m < R)

    {

        int tl = min(rsum[rt << 1], m - L + 1);

        int tr = min(lsum[rt << 1 | 1], R - m);

        ans = max(ans, tl + tr);

    }

    return ans;

}



int main()

{

    //freopen("input.txt", "r", stdin);



    int t, m, x, y;

    char op[5];

    RI(t);

    while(t--)

    {

        RII(n, m);

        num = 1;

        build(1, n, 1);

        while(m--)

        {

            RS(op);

            RII(x, y);

            if(op[0] == 'Q')PI(query(x + 1, y + 1, 1, n, 1));

            else update(x + 1, y, 1, n, 1);

        }

    }

    return 0;

}