Hdu3308LCIS线段树

 区间合并，和hotel那题差不多，只不过在合并区间的时候加和条件，变成左右儿子相连处是否上升，并且，对父亲的更新取得是最大值。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include<math.h>
using namespace std;

#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn = 111111;
int sum[maxn << 2], lsum[maxn << 2], rsum[maxn << 2];
int num[maxn];

void up(int l, int r, int rt)
{
    lsum[rt] = lsum[rt << 1];
    rsum[rt] = rsum[rt << 1 | 1];
    sum[rt] = max(sum[rt << 1], sum[rt << 1 | 1]);
    int mid = (l + r) >> 1;
    int m = r - l + 1;
    if (num[mid] < num[mid + 1]){
        if (lsum[rt] == (m - (m >> 1))) lsum[rt] += lsum[rt << 1 | 1];
        if (rsum[rt] == (m >> 1)) rsum[rt] += rsum[rt << 1];
        sum[rt] = max(sum[rt], lsum[rt << 1 | 1] + rsum[rt << 1]);
    }
}

void build(int l, int r, int rt)
{
    if (l == r){
        sum[rt] = lsum[rt] = rsum[rt] = 1;
        scanf("%d", &num[l]); return;
    }
    int mid = (l + r) >> 1;
    build(lson);
    build(rson);
    up(l, r, rt);
}

void update(int key, int add, int l, int r, int rt)
{
    if (l == r){
        num[l] = add; return;
    }
    int mid = (l + r) >> 1;
    if (key <= mid) update(key, add, lson);
    else update(key, add, rson);
    up(l, r, rt);
}

int ask(int L, int R, int l, int r, int rt)
{
    if (L <= l&&r <= R) return sum[rt];
    int mid = (l + r) >> 1;
    int ans = 0;
    if (L <= mid) ans = max(ans, ask(L, R, lson));
    if (R > mid) ans = max(ans, ask(L, R, rson));
    if (num[mid] < num[mid + 1]){
        int a = min(mid - L + 1, rsum[rt << 1]) + min(R - mid, lsum[rt << 1 | 1]);
        ans = max(ans, a);
    }
    return ans;
}

int main()
{
    char str[100];
    int Icase; int n, m, a, b;
    scanf("%d", &Icase);
    while (Icase--){
        scanf("%d%d", &n, &m);
        build(1, n, 1);
        for (int i = 0; i < m; i++){
            scanf("%s%d%d", str, &a, &b);
            if (str[0] == 'U'){
                a++;
                update(a, b, 1, n, 1);
            }
            else{
                a++; b++;
                cout << ask(a, b, 1, n, 1) << endl;
            }
        }
    }
    return 0;
}