HDU 3308 LCIS (线段树·单点更新·区间合并)

题意  给你一个数组  有更新值和查询两种操作  对于每次查询  输出对应区间的最长连续递增子序列的长度

基础的线段树区间合并  线段树维护三个值  对应区间的LCIS长度(lcis)  对应区间以左端点为起点的LCIS长度(lle)  对应区间以右端点为终点的LCIS长度(lri)  然后用val存储数组对应位置的值  当val[mid + 1] > val[mid] 的时候就要进行区间合并操作了

#include <cstdio>#include <algorithm>#define lc lp, s, mid#define rc rp, mid+1, e#define lp p<<1#define rp p<<1|1#define mid ((s+e)>>1)using namespace std;const int N = 100005, M = N * 4;int lcis[M], lle[M], lri[M], val[N];void pushup(int p, int s, int e){    lcis[p] = max(lcis[lp], lcis[rp]);    lle[p] = lle[lp], lri[p] = lri[rp];    if(val[mid + 1] > val[mid]) //合并条件    {        lcis[p] = max(lcis[p], lri[lp] + lle[rp]);        if(lle[lp] == mid + 1 - s) lle[p] += lle[rp];        if(lri[rp] == e - mid) lri[p] += lri[lp];    }}void build(int p, int s, int e){    if(s == e)    {        lcis[p] = lle[p] = lri[p] = 1;        scanf("%d", &val[s]);        return;    }    build(lc);    build(rc);    pushup(p, s, e);}void update(int p, int s, int e, int x, int v){    if(s == e)    {        val[s] = v;        return;    }    if(x <= mid) update(lc, x, v);    else if(x > mid) update(rc, x, v);    pushup(p, s, e);}int query(int p, int s, int e, int l, int r){    if(l <= s && e <= r) return lcis[p];    int ret = 0;    if(val[mid + 1] > val[mid])        ret = min(r, mid + lle[rp]) - max(l, mid + 1 - lri[lp]) + 1;    if(l <= mid)  ret = max(ret, query(lc, l, r));    if(r > mid)  ret = max(ret, query(rc, l, r));    return ret;}int main(){    int T, n, m, a, b;    char op[5];    scanf("%d", &T);    while(T--)    {        scanf("%d%d", &n, &m);        build(1, 0, n - 1);        while(m--)        {            scanf("%s%d%d", op, &a, &b);            if(op[0] == 'Q')                printf("%d\n", query(1, 0, n - 1, a, b));            else update(1, 0, n - 1, a, b);        }    }    return 0;}

LCIS

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

110 107 7 3 3 5 9 9 8 1 8 Q 6 6U 3 4Q 0 1Q 0 5Q 4 7Q 3 5Q 0 2Q 4 6U 6 10Q 0 9

 

Sample Output

11423125