杭电 3308 LCIS (线段树+单点更新+区间求和)

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 The next line has n integers(0<=val<=10 5). 
The next m lines each has an operation: 
U A B(0<=A,n , 0<=B=10 5) 
OR 
Q A B(0<=A<=B< n). 

Output

For each Q, output the answer.

Sample Input

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

Sample Output

1
1
4
2
3
1
2
5

思路：需要对最长递增子序列进行维护，每次一个个找肯定会爆。

AC代码：

#include
#include
using namespace std;

int n,m,num[100010];
struct node
{
	int l,r,c;//左右边界和区间长度 
	int ln,rn;//左右边界的值 
	int ls,rs,ms;//左右及区间的最长递增序列（LCIS） 
}tree[400010];

void up(int k)
{
	tree[k].ls=tree[k*2].ls;
	tree[k].rs=tree[k*2+1].rs;//更新左右边界值 
	tree[k].ln=tree[k*2].ln;
	tree[k].rn=tree[k*2+1].rn;//更新左右边界LCIS长度 
	tree[k].ms=max(tree[k*2].ms,tree[k*2+1].ms);//更新区间内LCIS长度
	if(tree[k*2].rn=l&&tree[k].r<=r) return tree[k].ms;
	int m=(tree[k].l+tree[k].r)/2,ans=0;
	if(l<=m)
	    ans = max(ans,query(l,r,2*k));
	if(r>m)
		ans=max(ans,query(l,r,2*k+1)); 
	if(tree[k*2].rn>1;
    if(t<=mid)
        insert(2*i,t,m);
    if(t>mid)
        insert(2*i+1,t,m);
    up(i);
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=1;i<=n;i++)  scanf("%d%*c",&num[i]);
		build(1,n,1);
		while(m--)
		{
			getchar();
			char c;
			int a,b;
			scanf("%c %d %d",&c,&a,&b);
			//cout<