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 | 1using 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;}